1.
What is the correct unit of work expressed in SI units?
Correct Answer
C. Kg m^2/s^2
Explanation
The correct unit of work expressed in SI units is kg m^2/s^2. Work is defined as the product of force and displacement, and the SI unit for force is kg m/s^2 (also known as a Newton). Therefore, when force is multiplied by displacement (in meters), the resulting unit for work is kg m^2/s^2.
2.
Can work be done on a system if there is no motion?
Correct Answer
D. No, because of the way work is defined.
Explanation
Work is defined as the transfer of energy that occurs when a force is applied to an object and the object is displaced in the direction of the force. If there is no motion, there is no displacement, and therefore no work is done. Work requires both force and displacement to be present in order to be defined and calculated.
3.
If you push twice as hard against a stationary brick wall, the amount of work you do
Correct Answer
D. Remains constant at zero.
Explanation
When you push against a stationary brick wall, the wall does not move. In order for work to be done, there must be a displacement in the direction of the force applied. Since the wall does not move, there is no displacement and therefore no work is done. Therefore, the amount of work remains constant at zero, regardless of how hard you push.
4.
A 50-N object was lifted 2.0 m vertically and is being held there. How much work is being done in holding the box in this position?
Correct Answer
D. 0 J
Explanation
When an object is held in a stationary position, no work is being done on the object. Work is defined as the transfer of energy that occurs when a force is applied over a distance. In this case, the object is not moving vertically, so there is no displacement and therefore no work is being done. Therefore, the correct answer is 0 J.
5.
If you walk 5.0 m horizontally forward at a constant velocity carrying a 10-N object, the amount of work you do is
Correct Answer
D. Zero.
Explanation
When an object is moved horizontally at a constant velocity, the work done on the object is zero. This is because work is defined as the product of force and displacement in the direction of the force. In this case, although there is a force (10 N) being applied, the displacement is in a perpendicular direction to the force (horizontal), resulting in no work being done. Therefore, the correct answer is zero.
6.
A container of water is lifted vertically 3.0 m then returned to its original position. If the total weight is 30 N, how much work was done?
Correct Answer
D. No work was done.
Explanation
Since the container is lifted vertically and then returned to its original position, the net displacement of the container is zero. Work is defined as the product of force and displacement, so if the displacement is zero, then no work is done. Therefore, no work was done in this scenario.
7.
Does the centripetal force acting on an object do work on the object?
Correct Answer
D. No, because the force and the displacement of the object are perpendicular.
Explanation
The centripetal force acting on an object does not do work on the object because the force and the displacement of the object are perpendicular. In order for work to be done, the force and the displacement must be in the same direction. Since the centripetal force always acts towards the center of the circular motion, while the displacement is tangential to the motion, they are perpendicular to each other and no work is done.
8.
You throw a ball straight up. Compare the sign of the work done by gravity while the ball goes up with the sign of the work done by gravity while it goes down.
Correct Answer
C. Work is - on the way up and + on the way down.
Explanation
When the ball is thrown straight up, the work done by gravity is negative because the force of gravity is acting in the opposite direction of the displacement of the ball. As the ball comes back down, the work done by gravity is positive because the force of gravity is now acting in the same direction as the displacement of the ball. Therefore, the correct answer is that the work is - on the way up and + on the way down.
9.
The area under the curve, on a Force versus position (F vs. x) graph, represents
Correct Answer
A. Work.
Explanation
The area under the curve on a Force versus position graph represents work. This is because work is defined as the product of force and displacement, and the area under the curve represents the integral of force with respect to displacement. Therefore, calculating the area under the curve gives us the amount of work done.
10.
On a plot of Force versus position (F vs. x), what represents the work done by the force F?
Correct Answer
C. The area under the curve
Explanation
The work done by a force is represented by the area under the curve on a plot of Force versus position (F vs. x). This is because the area under the curve represents the displacement of the object in the direction of the force. The work done is equal to the force applied multiplied by the distance moved in the direction of the force. Therefore, the area under the curve represents the total work done by the force.
11.
The quantity 1/2 mv^2 is
Correct Answer
A. The kinetic energy of the object.
Explanation
The quantity 1/2 mv^2 represents the kinetic energy of an object. Kinetic energy is the energy possessed by an object due to its motion. It is dependent on the mass of the object (m) and the square of its velocity (v^2). The formula 1/2 mv^2 is derived from the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Therefore, the correct answer is the kinetic energy of the object.
12.
If the net work done on an object is positive, then the object's kinetic energy
Correct Answer
C. Increases.
Explanation
When the net work done on an object is positive, it means that the total work done on the object is in the same direction as its motion. This indicates that the object is gaining energy and its kinetic energy is increasing. Therefore, the correct answer is "increases."
13.
If the net work done on an object is negative, then the object's kinetic energy
Correct Answer
A. Decreases.
Explanation
When the net work done on an object is negative, it means that the work done on the object is in the opposite direction of its motion. This implies that the object is losing energy, resulting in a decrease in its kinetic energy. Therefore, the correct answer is that the object's kinetic energy decreases.
14.
If the net work done on an object is zero, then the object's kinetic energy
Correct Answer
B. Remains the same.
Explanation
If the net work done on an object is zero, it means that the total amount of work done on the object is equal to zero. This implies that there is no change in the object's kinetic energy. Therefore, the object's kinetic energy remains the same.
15.
A truck weighs twice as much as a car, and is moving at twice the speed of the car. Which statement is true about the truck's kinetic energy compared to that of the car?
Correct Answer
D. The truck has 8 times the kinetic energy of the car.
Explanation
The kinetic energy of an object is directly proportional to its mass and the square of its velocity. In this scenario, the truck weighs twice as much as the car and is moving at twice the speed of the car. Since the kinetic energy is proportional to the square of the velocity, the truck's kinetic energy is 4 times that of the car. Additionally, since the truck weighs twice as much as the car, its kinetic energy is also 4 times that of the car. Therefore, the truck has a total of 4 times 4, which is 8 times, the kinetic energy of the car.
16.
Car J moves twice as fast as car K, and car J has half the mass of car K. The kinetic energy of car J, compared to car K is
Correct Answer
B. 2 to 1.
Explanation
Car J moves twice as fast as car K, which means that its velocity is twice that of car K. The kinetic energy of an object is proportional to the square of its velocity. Since car J has twice the velocity of car K, its kinetic energy will be four times greater. However, car J has half the mass of car K. The kinetic energy is also proportional to the mass of the object. Since car J has half the mass of car K, its kinetic energy will be half that of car K. Therefore, the ratio of the kinetic energy of car J to car K is 2 to 1.
17.
An object hits a wall and bounces back with half of its original speed. What is the ratio of the final kinetic energy to the initial kinetic energy?
Correct Answer
B. 1/4
Explanation
When an object hits a wall and bounces back with half of its original speed, its final kinetic energy is proportional to the square of its final speed. Since the final speed is half of the initial speed, the final kinetic energy will be one-fourth (1/4) of the initial kinetic energy. Therefore, the ratio of the final kinetic energy to the initial kinetic energy is 1/4.
18.
A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass?
Correct Answer
A. 25 to 9
Explanation
The ratio of the brick's mass to the rock's mass is 25 to 9 because kinetic energy is directly proportional to mass. Since both objects have the same kinetic energy, the ratio of their masses will be the same as the ratio of their speeds squared. The ratio of the speeds squared is (3^2)/(5^2) = 9/25, so the ratio of the masses is 25 to 9.
19.
A 4.0-kg mass is moving with speed 2.0 m/s. A 1.0-kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?
Correct Answer
C. Both travel the same distance.
Explanation
Both objects will travel the same distance before stopping because the distance traveled is determined by the work done on the object, which is equal to the force applied multiplied by the distance traveled. In this case, both objects experience the same braking force, so the work done on each object is the same. Therefore, both objects will travel the same distance before coming to a stop.
20.
You slam on the brakes of your car in a panic, and skid a certain distance on a straight, level road. If you had been traveling twice as fast, what distance would the car have skidded, under the same conditions?
Correct Answer
A. It would have skidded 4 times farther.
Explanation
If the car had been traveling twice as fast, it would have had twice the kinetic energy. When the brakes are applied, the car's kinetic energy is converted into heat energy through friction, causing the car to skid. Since the car has twice the initial kinetic energy, it would take four times as much energy to bring the car to a stop, resulting in the car skidding four times farther.
21.
A planet of constant mass orbits the Sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet's kinetic energy?
Correct Answer
D. It increases when the planet approaches the Sun, and decreases when it moves farther
away.
Explanation
As the planet moves closer to the Sun in its elliptical orbit, it experiences a stronger gravitational force. This increased force causes the planet to accelerate, resulting in an increase in its kinetic energy. Conversely, as the planet moves farther away from the Sun, the gravitational force weakens, causing the planet to decelerate and its kinetic energy to decrease. Therefore, the planet's kinetic energy increases when it approaches the Sun and decreases when it moves farther away.
22.
The quantity mgy is
Correct Answer
B. The gravitational potential energy of the object.
Explanation
The quantity mgy represents the product of the mass (m), the acceleration due to gravity (g), and the height (y) of the object. This equation is derived from the formula for gravitational potential energy, which is given by mgh. The only difference is that y is used instead of h to represent the height. Therefore, mgy represents the gravitational potential energy of the object.
23.
The quantity 1/2 kx^2 is
Correct Answer
B. The elastic potential energy of the object.
Explanation
The quantity 1/2 kx^2 represents the elastic potential energy of the object. Elastic potential energy is the energy stored in a stretched or compressed object, and it is directly proportional to the square of the displacement (x) from the equilibrium position and the spring constant (k). Therefore, the given expression, 1/2 kx^2, accurately represents the elastic potential energy of the object.
24.
Is it possible for a system to have negative potential energy?
Correct Answer
B. Yes, since the choice of the zero of potential energy is arbitrary.
Explanation
The correct answer is "Yes, since the choice of the zero of potential energy is arbitrary." This means that the reference point for measuring potential energy can be chosen arbitrarily, and therefore it is possible for a system to have negative potential energy depending on the chosen reference point. The total energy of the system can still be positive, even if the potential energy is negative.
25.
An object is released from rest a height h above the ground. A second object with four times the mass of the first if released from the same height. The potential energy of the second object compared to the first is
Correct Answer
D. Four times as much.
Explanation
When an object is released from rest, its potential energy is converted into kinetic energy as it falls. The potential energy of an object is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the height is the same for both objects, but the mass of the second object is four times greater than the mass of the first object. Therefore, the potential energy of the second object is four times greater than the potential energy of the first object.
26.
A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, the potential energy of the spring will be
Correct Answer
D. 4 times as much.
Explanation
When another 0.200-kg mass is added to the spring, the total mass attached to the spring becomes 0.400 kg. The potential energy of a spring is given by the equation PE = (1/2)kx², where k is the spring constant and x is the displacement from the equilibrium position. Since the displacement remains the same at 5.0 cm, and the mass is doubled, the potential energy of the spring will be four times as much. This is because the potential energy is directly proportional to the square of the displacement and the mass.
27.
The total mechanical energy of a system
Correct Answer
D. Is constant, only if conservative forces act.
Explanation
The correct answer is that the total mechanical energy of a system is constant, only if conservative forces act. This means that if the only forces acting on a system are conservative forces (such as gravity or springs), then the total mechanical energy (the sum of kinetic and potential energy) will remain constant over time. Non-conservative forces, such as friction or air resistance, can convert mechanical energy into other forms (such as heat or sound), causing the total mechanical energy of the system to change.
28.
An acorn falls from a tree. Compare its kinetic energy K, to its potential energy U.
Correct Answer
A. K increases and U decreases.
Explanation
As the acorn falls from the tree, its potential energy is converted into kinetic energy. The potential energy of the acorn decreases as it falls closer to the ground, while its kinetic energy increases due to its motion. Therefore, the correct answer is that the kinetic energy increases and the potential energy decreases.
29.
Describe the energy of a car driving up a hill.
Correct Answer
C. Both kinetic and potential
Explanation
When a car is driving up a hill, it possesses both kinetic and potential energy. Kinetic energy is the energy of motion, which the car has as it moves up the hill. Potential energy is the energy that an object possesses due to its position or height above the ground. As the car gains height while driving up the hill, it gains potential energy. Therefore, the energy of a car driving up a hill is both kinetic and potential.
30.
A lightweight object and a very heavy object are sliding with equal speeds along a level frictionless surface. They both slide up the same frictionless hill. Which rises to a greater height?
Correct Answer
C. They both slide to the same height.
Explanation
The correct answer is that they both slide to the same height. Although the heavy object has greater kinetic energy, this does not affect the height it reaches on the hill. The height reached by an object sliding up a hill is determined by its initial speed and the shape of the hill, not its mass or kinetic energy. Therefore, both objects will reach the same height on the hill regardless of their weights or speeds.
31.
Consider two masses m(1) and m(2) at the top of two frictionless inclined planes. Both masses start from rest at the same height. However, the plane on which m(1) sits is at an angle of 30° with the horizontal, while the plane on which m(2) sits is at 60°. If the masses are released, which is going faster at the bottom of its plane?
Correct Answer
C. They both are going the same speed.
Explanation
Both masses m(1) and m(2) are released from the same height and experience the same gravitational force. The speed of an object sliding down an inclined plane depends only on the height from which it is released and not on the angle of the incline. Therefore, both masses will have the same speed at the bottom of their respective planes. The masses and angles of the inclines do not affect the speed.
32.
A ball falls from the top of a building, through the air (air friction is present), to the ground below. How does the kinetic energy (K) just before striking the ground compare to the potential energy (U) at the top of the building?
Correct Answer
C. K is less than U.
Explanation
The kinetic energy (K) just before striking the ground is less than the potential energy (U) at the top of the building. This is because as the ball falls, some of its potential energy is converted into kinetic energy due to the force of gravity. However, air friction acts as a resistance force, causing some of the energy to be lost as heat. Therefore, the ball loses some of its potential energy and does not gain an equal amount of kinetic energy, resulting in K being less than U.
33.
A ball drops some distance and gains 30 J of kinetic energy. Do not ignore air resistance. How much gravitational potential energy did the ball lose?
Correct Answer
A. More than 30 J
Explanation
When a ball drops and gains kinetic energy, it means that it is converting gravitational potential energy into kinetic energy. Since the ball gains 30 J of kinetic energy, it must have lost more than 30 J of gravitational potential energy.
34.
A ball drops some distance and loses 30 J of gravitational potential energy. Do not ignore air resistance. How much kinetic energy did the ball gain?
Correct Answer
C. Less than 30 J
Explanation
When a ball drops and loses gravitational potential energy, it gains an equal amount of kinetic energy according to the law of conservation of energy. Since the ball loses 30 J of gravitational potential energy, it gains 30 J of kinetic energy. Therefore, the correct answer is "exactly 30 J" and not "less than 30 J".
35.
The quantity Fd/t is
Correct Answer
D. The power supplied to the object by the force.
Explanation
The quantity Fd/t represents the power supplied to the object by the force. Power is defined as the rate at which work is done or the amount of energy transferred per unit time. In this case, F represents the force applied to the object, d represents the displacement of the object, and t represents the time taken. Therefore, Fd/t represents the power supplied to the object by the force.
36.
What is the correct unit of power expressed in SI units?
Correct Answer
C. Kg m^2/s^3
Explanation
The correct unit of power expressed in SI units is kg m^2/s^3. This unit represents the amount of work done or energy transferred per unit time. It is derived from the basic SI units of kilogram (kg) for mass, meter (m) for distance, and second (s) for time. The exponent of 2 for the meter term indicates that power is proportional to the square of the distance, while the exponent of 3 for the second term indicates that power is inversely proportional to the cube of time.
37.
Of the following, which is not a unit of power?
Correct Answer
A. Watt/second
Explanation
The correct answer is watt/second. This is because watt/second is not a unit of power. Power is measured in watts, which is a unit of energy transfer per unit time. The other options, newton-meter/second and joule/second, are not units of power either. Newton-meter/second is a unit of torque, and joule/second is a unit of energy transfer rate.
38.
Compared to yesterday, you did 3 times the work in one-third the time. To do so, your power output must have been
Correct Answer
D. 9 times yesterday's power output.
Explanation
If you did 3 times the work in one-third the time compared to yesterday, it means that your power output increased. To do 3 times the work in one-third the time, you would need to exert 9 times more power than yesterday.
39.
To accelerate your car at a constant acceleration, the car's engine must
Correct Answer
C. Develop ever-increasing power.
Explanation
In order to accelerate a car at a constant acceleration, the car's engine must develop ever-increasing power. This is because as the car accelerates, it requires more power to overcome the increasing resistance and maintain the constant acceleration. If the engine maintained a constant power output or developed ever-decreasing power, the car would not be able to accelerate at a constant rate. Additionally, maintaining a constant turning speed is not relevant to accelerating the car.
40.
An object is lifted vertically 2.0 m and held there. If the object weighs 90 N, how much work was done in lifting it?
Correct Answer
B. 180 J
Explanation
When an object is lifted vertically, work is done against the force of gravity. The work done is equal to the force applied multiplied by the distance over which the force is applied. In this case, the object weighs 90 N and is lifted vertically 2.0 m. Therefore, the work done in lifting the object is 90 N * 2.0 m = 180 J.
41.
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity of 0.50 m/s. What is the work done by the weight of the book?
Correct Answer
B. -10 J
Explanation
The work done by the weight of the book is calculated using the formula: work = force x distance x cos(theta). In this case, the force is the weight of the book, which is 10 N. The distance is 1.0 m. Since the book is being lifted upwards, the angle between the force and displacement is 180 degrees, so cos(theta) = -1. Therefore, the work done is -10 J.
42.
A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the work done by the 5500 N force?
Correct Answer
A. 2.75 * 10^5 J
Explanation
The work done by a force is calculated using the formula: work = force * distance. In this case, the force is 5500 N and the distance is 50.0 m. Plugging in these values into the formula, we get: work = 5500 N * 50.0 m = 275,000 J. Therefore, the work done by the 5500 N force is 2.75 * 10^5 J.
43.
A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the work done by the weight of the elevator?
Correct Answer
B. -2.45 * 10^5 J
Explanation
The work done by the weight of the elevator can be calculated using the formula: work = force * distance. In this case, the force is the weight of the elevator, which is equal to its mass multiplied by the acceleration due to gravity. Since the elevator is pulled upward, the force of gravity is acting in the opposite direction, so the work done by the weight of the elevator is negative. Therefore, the correct answer is -2.45 * 10^5 J.
44.
A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the net work done on the elevator?
Correct Answer
C. 3.00 * 10^4 J
Explanation
The net work done on the elevator can be calculated using the formula W = F * d, where W is the work done, F is the force applied, and d is the distance traveled. In this case, the force applied is 5500 N and the distance traveled is 50.0 m. Multiplying these values together gives a net work done of 275,000 J or 2.75 * 10^5 J. Therefore, the correct answer is 2.75 * 10^5 J.
45.
A 30-N box is pulled 6.0 m up along a 37° inclined plane. What is the work done by the weight (gravitational force) of the box?
Correct Answer
B. -1.1 * 10^2J
Explanation
The work done by the weight (gravitational force) of the box can be calculated using the formula: Work = Force x Distance x cos(theta), where theta is the angle between the force and the displacement. In this case, the force is the weight of the box, which is given as 30 N. The distance is 6.0 m and the angle is 37°. Plugging these values into the formula, we get: Work = 30 N x 6.0 m x cos(37°) = -1.1 * 10^2 J. Therefore, the correct answer is -1.1 * 10^2 J.
46.
A 4.00-kg box of fruit slides 8.0 m down a ramp, inclined at 30.0° from the horizontal. If the box slides at a constant velocity of 5.00 m/s, what is the work done by the weight of the box?
Correct Answer
A. 157 J
Explanation
The work done by the weight of the box can be calculated using the formula: work = force x distance x cos(angle). In this case, the force is the weight of the box, which can be calculated using the formula: weight = mass x gravity. The distance is given as 8.0 m and the angle is 30.0 degrees. Plugging in the values, we can calculate the work done by the weight of the box to be 157 J.
47.
Matthew pulls his little sister Sarah in a sled on an icy surface (assume no friction), with a force of 60.0 N at an angle of 37.0° pward from the horizontal. If he pulls her a distance of 12.0 m, what is the work done by Matthew?
Correct Answer
C. 575 J
Explanation
Matthew is pulling the sled with a force of 60.0 N at an angle of 37.0° upward from the horizontal. The work done by Matthew can be calculated using the formula: work = force * distance * cos(angle). Plugging in the given values, we get: work = 60.0 N * 12.0 m * cos(37.0°) = 575 J. Therefore, the work done by Matthew is 575 J.
48.
A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final kinetic energy of the cart?
Correct Answer
B. 2.0 * 10^3 J
Explanation
The final kinetic energy of the cart can be calculated using the formula K.E. = 1/2 * m * v^2, where m is the mass of the cart and v is the final velocity. In this case, the force applied is causing the cart to accelerate, so we need to find the final velocity first. We can use Newton's second law of motion, F = m * a, where F is the force applied, m is the mass of the cart, and a is the acceleration. Rearranging the equation, we get a = F/m. Plugging in the values, we have a = 200 N / 55 kg = 3.64 m/s^2. Now, we can use the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity (which is 0 m/s since the cart is initially at rest), a is the acceleration, and s is the distance traveled. Plugging in the values, we have v^2 = 0 + 2 * 3.64 m/s^2 * 10 m = 72.8 m^2/s^2. Taking the square root of both sides, we get v = 8.53 m/s. Now, we can calculate the final kinetic energy using the formula K.E. = 1/2 * m * v^2. Plugging in the values, we have K.E. = 1/2 * 55 kg * (8.53 m/s)^2 = 2.0 * 10^3 J. Therefore, the correct answer is 2.0 * 10^3 J.
49.
A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart?
Correct Answer
C. 8.5 m/s
Explanation
The final speed of the cart can be determined using Newton's second law of motion, which states that force is equal to mass times acceleration. In this case, the force applied is 200 N and the mass of the cart is 55 kg. The acceleration can be calculated by dividing the force by the mass, which gives us 200 N / 55 kg = 3.64 m/s^2. To find the final speed, we can use the equation v^2 = u^2 + 2as, where u is the initial velocity (0 m/s), a is the acceleration (3.64 m/s^2), and s is the distance (10 m). Plugging in the values, we get v^2 = 0^2 + 2(3.64 m/s^2)(10 m) = 72.8 m^2/s^2. Taking the square root of both sides, we get v = √72.8 m^2/s^2 = 8.5 m/s. Therefore, the final speed of the cart is 8.5 m/s.
50.
A 10-kg mass is moving with a speed of 5.0 m/s. How much work is required to stop the mass?
Correct Answer
D. 125 J
Explanation
To stop the mass, work must be done to bring it to a complete stop. The work done is equal to the change in kinetic energy. The initial kinetic energy of the mass is given by 1/2 * mass * velocity squared, which is equal to 1/2 * 10 kg * (5.0 m/s)^2 = 125 J. Therefore, 125 J of work is required to stop the mass.