# Unit 7 - Forces

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MrCanning
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Quizzes Created: 3 | Total Attempts: 1,792
Questions: 15 | Attempts: 199  Settings  A 15 question quiz on forces.

• 1.

### Which forces could be labelled on a yo-yo?

• A.

Gravity

• B.

Normal

• C.

Friction

• D.

Centripetal

A. Gravity
C. Friction
D. Centripetal
Explanation
The forces that could be labeled on a yo-yo are gravity, friction, and centripetal. Gravity is the force that pulls the yo-yo down towards the ground. Friction is the force that opposes the motion of the yo-yo, allowing it to spin and stay in place. Centripetal force is the force that keeps the yo-yo moving in a circular path when it is spinning. These forces work together to control the motion and behavior of the yo-yo.

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• 2.

### What is the symbol for acceleration due to gravity?

• A.

A

• B.

G

• C.

9.8

• D.

M/s^2

B. G
Explanation
The symbol for acceleration due to gravity is "g". This symbol is commonly used in physics to represent the acceleration experienced by an object due to the force of gravity. It is a constant value on Earth and is approximately equal to 9.8 m/s^2.

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• 3.

### A body weighs 125 N in a place where acceleration due to gravity is 10 m / s2. Calculated: a) the mass of the object, b) the weight of the object in a place where the acceleration of gravity is 9.65 m / s2.

• A.

1250 kg and 121 N

• B.

1250 g and 12.1 N

• C.

12.5 kg and 121 N

• D.

12.5 kg and 12.1 N

C. 12.5 kg and 121 N
Explanation
The weight of an object is calculated by multiplying its mass by the acceleration due to gravity. In this case, the weight is given as 125 N and the acceleration due to gravity is given as 10 m/s^2. By rearranging the formula, we can solve for the mass of the object. Mass = Weight / Acceleration due to gravity. Therefore, the mass of the object is 125 N / 10 m/s^2 = 12.5 kg.

To calculate the weight of the object in a place where the acceleration of gravity is 9.65 m/s^2, we use the same formula. Mass = Weight / Acceleration due to gravity. Plugging in the values, we get Weight = Mass * Acceleration due to gravity = 12.5 kg * 9.65 m/s^2 = 121 N. Therefore, the weight of the object in that place is 121 N.

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• 4.

### A 28-ton truck changes its speed over 200 s from 36 km/h to 54 km/h. Determine the force exerted by it.

• A.

1400 N

• B.

500 N

• C.

900 N

• D.

700 N

D. 700 N
Explanation
To determine the force exerted by the truck, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. Since the mass of the truck is given as 28 tons, we need to convert it to kilograms by multiplying it by 1000 (1 ton = 1000 kg). The change in speed can be calculated by subtracting the initial speed from the final speed. The time taken for the change in speed is given as 200 s. Dividing the change in speed by the time gives us the acceleration. Finally, multiplying the mass by the acceleration gives us the force exerted by the truck, which is 700 N.

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• 5.

### On a body of 200 kg, a force is exerted 400 N. Calculate the acceleration acquired by the body and use this to, assuming the object begins at rest, how fast will be travelling(in m/s) after 10 s?(Type only the number e.g. 15)

20
Explanation
The acceleration of an object can be calculated using the formula F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration. In this case, the force applied is 400 N and the mass of the object is 200 kg. Plugging these values into the formula, we get 400 = 200a. Solving for a, we find that the acceleration is 2 m/s^2. To find the velocity of the object after 10 seconds, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since the object starts at rest), a is the acceleration, and t is the time. Plugging in the values, we get v = 0 + 2(10) = 20 m/s.

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• 6.

### Which units could be used for pressure? (Write down the formula first to help)

• A.

N/m2

• B.

Pa

• C.

KPa

• D.

N/mm2

A. N/m2
B. Pa
C. KPa
D. N/mm2
Explanation
The units N/m2, Pa, kPa, and N/mm2 could be used for pressure. Pressure is defined as force per unit area, so the formula for pressure is force divided by area. The unit N/m2 represents Newton per square meter, which is the SI unit for pressure. Pa stands for Pascal, which is also a unit of pressure in the SI system. kPa stands for kilopascal, which is equal to 1000 pascals. N/mm2 represents Newton per square millimeter, which is another unit for pressure.

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• 7.

### What is the pressure produced by a force of 25000 N acting on an area of 5 m2.

• A.

5 Pa

• B.

50 Pa

• C.

500 Pa

• D.

5000 Pa

D. 5000 Pa
Explanation
The pressure produced by a force is calculated by dividing the force by the area on which it is acting. In this case, the force is 25000 N and the area is 5 m2. Therefore, the pressure is 25000 N / 5 m2 = 5000 Pa.

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• 8.

### Determine the surface, in m2, of the base of a monument of 30 tonnes if the pressure exerted on its base is 25 N/cm2. (g=10 m/s2)

• A.

0.6

• B.

0.8

• C.

1.0

• D.

1.2

D. 1.2
Explanation
The pressure exerted on the base of the monument can be calculated using the formula pressure = force/area. We are given the pressure (25 N/cm2) and the force (30 tonnes = 30000 kg x 10 m/s2 = 300000 N). Rearranging the formula, we can solve for the area: area = force/pressure. Plugging in the values, we get area = 300000 N / 25 N/cm2 = 12000 cm2. To convert this to m2, we divide by 10000 (since 1 m2 = 10000 cm2). Therefore, the surface area of the base of the monument is 1.2 m2.

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• 9.

### Calculate the pressure of a monument of 30 tonnes if its base is a rectangle measuring 20 cm long and 1.5 m wide. (g = 10 m / s2)

• A.

1000000 Pa

• B.

500 Pa

• C.

20000000000000 Pa

• D.

150 Pa

A. 1000000 Pa
Explanation
The pressure of an object is calculated by dividing the force acting on the object by the area over which the force is applied. In this case, the force acting on the monument is its weight, which is given as 30 tonnes. To convert this to Newtons, we multiply by the acceleration due to gravity (g = 10 m/s^2). The weight of the monument is therefore 30,000 kg * 10 m/s^2 = 300,000 N.

The area over which this force is applied is the base of the monument, which is a rectangle measuring 20 cm (or 0.2 m) long and 1.5 m wide. The area is therefore 0.2 m * 1.5 m = 0.3 m^2.

Finally, we can calculate the pressure by dividing the force (300,000 N) by the area (0.3 m^2), which gives us 1,000,000 Pa.

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• 10.

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• 11.

### Calculate the pressure in the S.I units, which a diver feels when they are 5 m deep. Density of seawater is 1.1 g / cm3. (g=10 m/s2)

• A.

50000 Pa

• B.

55000 Pa

• C.

60000 Pa

• D.

65000 Pa

B. 55000 Pa
Explanation
The pressure experienced by a diver at a certain depth is given by the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the density of seawater is given as 1.1 g/cm3, which can be converted to kg/m3 by dividing by 1000. The acceleration due to gravity is 10 m/s2. The depth is given as 5 m. Plugging these values into the formula, we get P = (1.1 kg/m3)(10 m/s2)(5 m) = 55 Pa. Therefore, the correct answer is 55000 Pa.

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• 12.

### A child gets into a pool, dislodging water weighing 600 N.What mass is the boy assuming he is only just floating in the water? (Use g=9.8)

• A.

61 kg

• B.

68 kg

• C.

74 kg

• D.

80 kg

A. 61 kg
Explanation
The boy is assuming a mass of 61 kg because when he gets into the pool, he displaces water weighing 600 N. According to Archimedes' principle, the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. Since the buoyant force is equal to the weight of the boy, and the weight of the water displaced is 600 N, the boy's weight must also be 600 N. Using the equation F = mg, where F is the weight, m is the mass, and g is the acceleration due to gravity, we can solve for the mass, which is 61 kg.

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• 13.

### A steel ball of radius 5 cm is immersed in pure water. Calculate the thrust of the water and the weight of the ball (density of steel = 7.9 g / cm3). Vsphere = (4/3)·r3

• A.

2.63 and 18.9

• B.

1.63 and 12.9

• C.

5.63 and 14.0

• D.

3.63 and 3.9 Back to top