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Mechanics - Laws of Motion - SS2 Physics Past Questions and Answers - page 2

11

A 5 kg object experiences a net force of 20 N. What is the acceleration of the object?

According to Newton's Second Law (F = ma), the acceleration of an object is given by a = F/m. Plugging in the values, we get a = 20 N / 5 kg = 4 m/s2.

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12

An object of mass 2 kg is pushed with a force of 10 N. What is the acceleration of the object?

Using Newton's Second Law (F = ma), we rearrange the equation to solve for acceleration: a = F/m. Plugging in the values, we get a = 10 N / 2 kg = 5 m/s2.

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13

A car of mass 1000 kg accelerates from rest to a speed of 20 m/s in 10 seconds. What is the average force exerted on the car?

We can use Newton's Second Law (F = ma) to find the force. The acceleration is given by a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time. Plugging in the values, we get a = (20 m/s - 0 m/s) / 10 s = 2 m/s2. Then, F = ma = 1000 kg x 2 m/s2 = 2000 N.

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14

A force of 50 N is applied to an object of mass 10 kg. What is the acceleration of the object?

Using Newton's Second Law (F = ma), we rearrange the equation to solve for acceleration: a = F/m. Plugging in the values, we get a = 50 N / 10 kg = 5 m/s2.

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15

A ball is thrown vertically upwards with an initial velocity of 20 m/s. How high does the ball rise before coming back down? (Take g = 10 m/s2)

When the ball reaches its highest point, its velocity becomes 0 m/s. Using the equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, -g), and s is the displacement, we can solve for the displacement. Plugging in the values, we get 02 = 202 + 2(-10)s. Solving for s, we get s = 100 m.

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16

The force of friction between two surfaces depends on:

A

The mass of the objects in contact

 

B

The area of contact between the surfaces

 

C

The coefficient of friction between the surfaces

 

D

Both b) and c)

correct option: d

Explanation: The force of friction is influenced by the area of contact between the surfaces and the coefficient of friction, which is a property of the materials in contact.

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17

The coefficient of friction is defined as:

A

The ratio of the frictional force to the applied force

 

B

The ratio of the normal force to the frictional force

 

C

The ratio of the frictional force to the weight of the object

 

D

The ratio of the applied force to the normal force

correct option: c

Explanation: The coefficient of friction (μ) is defined as the ratio of the frictional force (F) to the weight of the object (W). μ = F/W.

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18

Which type of friction is generally larger?

A

Static friction

 

B

Kinetic friction

 

C

Rolling friction

 

 

D

Fluid friction

correct option: a

Explanation: Static friction is generally larger than kinetic friction. It is the force that opposes the motion of an object at rest, and it must be overcome to set an object in motion.

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19

The coefficient of static friction is always:

A

Greater than the coefficient of kinetic friction

 

B

Equal to the coefficient of kinetic friction

 

 

C

Smaller than the coefficient of kinetic friction

 

D

Unrelated to the coefficient of kinetic friction

correct option: a

Explanation: The coefficient of static friction (μs) is generally greater than the coefficient of kinetic friction (μk) for a given pair of surfaces. The static friction is higher because it needs to be overcome to initiate motion.

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20

The force of friction is directly proportional to:

A

The mass of the object

 

 

B

The normal force between the surfaces

 

C

The velocity of the object

 

D

The coefficient of restitution

correct option: b

Explanation: The force of friction is directly proportional to the normal force between the surfaces. The normal force is the force exerted perpendicular to the surfaces in contact.

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