# Work And Power Quiz

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Kevin Shaw
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Quizzes Created: 7 | Total Attempts: 5,036
Questions: 10 | Attempts: 520

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• 1.

### A piston, moving through a distance of 15 cm, pushes a box weighing 8.0 kg onto a conveyor belt with a force of 40 N. How much work is done by the piston on the box? A   6.0 J          B   120 J          C   320 J          D   600 J

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A.
Explanation
The work done by the piston on the box can be calculated using the formula: work = force x distance. In this case, the force applied by the piston is 40 N and the distance moved by the piston is 15 cm (which is equivalent to 0.15 m). Therefore, the work done by the piston on the box is 40 N x 0.15 m = 6.0 J.

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• 2.

### How high can a worker lift a 40.00-kg bag of sand if he produces 4,000 J of energy? Assume no energy is used to overcome friction.  A   1.020 m          B   10.20 m          C   102.0 m          D   1,020 m

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B.
Explanation
The correct answer is A. 1.020 m. The height that a worker can lift a bag of sand is determined by the amount of potential energy they can produce. The potential energy is equal to the mass of the object multiplied by the acceleration due to gravity multiplied by the height. In this case, the potential energy produced is 4,000 J. By rearranging the equation and substituting the values, we can solve for the height, which is approximately 1.020 m.

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• 3.

### Neglecting friction, if a child exerts a force of 85 N on the handle of a wagon that makes a 35Â° angle with the horizontal, how far is the wagon pulled when 280 J of work are done?                                                                                                                           A    â‰¤ 5.0 m B    > 5.0 m but â‰¤ 10 m C    > 10 m but â‰¤ 15 m D    > 15 m but â‰¤ 20 m

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A.
Explanation
The work done on an object is equal to the force applied on the object multiplied by the distance the object is moved in the direction of the force. In this case, the force applied by the child is 85 N and the work done is 280 J. To find the distance the wagon is pulled, we can rearrange the equation for work to solve for distance: distance = work / force. Plugging in the values, we get distance = 280 J / 85 N = 3.29 m. Therefore, the wagon is pulled a distance of approximately 3.29 meters.

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• 4.

### This is a graph representing work versus time. What does the slope of the graph represent? A   accelerationB   impulseC   powerD   velocity

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C.
Explanation
The slope of the graph represents velocity. This is because velocity is defined as the rate of change of displacement with respect to time, and the slope of a graph represents the rate of change of the dependent variable (in this case, work) with respect to the independent variable (time). Therefore, the slope of the graph represents the rate at which work is being done, which is equivalent to velocity.

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• 5.

### If a forklift raises a 76-kg load a distance of 2.5 m, how much work has it done? A   80 J B  19 JC   300 JD   1900 J

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Explanation
The correct answer is C. 300 J.

Work is calculated by multiplying the force applied by the distance over which it is applied. In this case, the force applied by the forklift is the weight of the load, which is the mass of the load multiplied by the acceleration due to gravity (9.8 m/s^2). So, the force is 76 kg * 9.8 m/s^2 = 744.8 N.

The distance over which the force is applied is 2.5 m.

Therefore, the work done by the forklift is 744.8 N * 2.5 m = 1862 J.

However, since the question asks for the work done in joules, the answer should be rounded to the nearest whole number, which is 1900 J.

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• 6.

### The amount of power required to move an object can be increased without changing the amount of work required.  How can this happen?   A increase the time required to do the work B increase the friction on the surface over which the object is moving C increase the weight of the object being moved D decrease the time required to do the work

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D.
Explanation
B increase the friction on the surface over which the object is moving

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• 7.

### A woman driving a 2000-kg car at 15 m/s fully applies the brakes 50 m from a stoplight. If the car stops 5.0 m before the light, what is the magnitude of the average force applied by the brakes?  A   4.1 E 103 N B   4.5 E 103 N C   5.0 E 103 N D   4.5 E 104 N

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Explanation
The correct answer is C. The magnitude of the average force applied by the brakes is 5.0 E 103 N. This can be determined using the equation F = m*a, where F is the force, m is the mass of the car, and a is the acceleration. The acceleration can be calculated using the equation a = (vf^2 - vi^2) / (2*d), where vf is the final velocity (0 m/s in this case), vi is the initial velocity (15 m/s in this case), and d is the distance (50 m - 5 m = 45 m in this case). Plugging in the values, we get a = (0^2 - 15^2) / (2*45) = -225 / 90 = -2.5 m/s^2. Finally, plugging the values of m and a into the equation F = m*a, we get F = 2000 kg * (-2.5 m/s^2) = -5000 N. Since the force is negative, we take the magnitude, which is 5000 N, or 5.0 E 103 N.

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• 8.

### A boy pushes his wagon at constant speed along a level sidewalk. The graph below represents the relationship between the horizontal force exerted by the boy and the distance the wagon moves. What is the total work done by the boy in pushing the wagon 4.0 meters? A   5.0 J          B   120 J          C   7.5 J          D   180 J

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B.
Explanation
The correct answer is B 120 J. This can be determined by calculating the area under the graph, which represents the work done. The area under the graph between 0 and 4.0 meters is a rectangle with a base of 4.0 meters and a height of 30 N. The formula for calculating work is work = force x distance, so the work done is 30 N x 4.0 meters = 120 J.

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• 9.

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B.
• 10.

### A 60.0-kilogram runner has 1920 joules of kinetic energy. At what speed is she running? A  5.66 m/s     B  32.0 m/s     C  8.00 m/s     D  64.0 m/s

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C.
Explanation
The correct answer is A, 5.66 m/s. This can be determined using the formula for kinetic energy, which is KE = 1/2 * mass * velocity^2. Rearranging the formula to solve for velocity, we get velocity = sqrt(2 * KE / mass). Plugging in the given values, we get velocity = sqrt(2 * 1920 / 60) = 5.66 m/s.

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• Current Version
• Mar 18, 2023
Quiz Edited by
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• Nov 01, 2010
Quiz Created by
Kevin Shaw

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