NEET Laws of Motion MCQs & PYQs

Q: A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Maximum Friction force action on an object is μN. Here μ=0.2 and Normal reaction force will be equal to applied force i.e 10 N., So, friction force is 0.2*10 = 2N

Practice NEET Laws of Motion Questions

Question 11:

easy

A body of weight W1 is suspended from the ceiling of a room through a chain of weight W2. The ceiling pulls the chain by a force

Total weight of chain + block system is W1 +W2.

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Question 12:

easy

When a constant force is applied to a body, it moves with uniform

From Newton's second law of motion, F= m.a

if force is constant acceleration remains constant.

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Question 13:

easy

A particle is moving with a constant speed along a straight line path. A force is not required to

As the particle moves with constant speed in straight line its velocity is constant i.e accleration is zero. From Newton's second law of motion

F=m.a

Force required to keep moving with uniform velocity is zero.

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Question 14:

difficult

A block is kept on the frictionless inclined surface with angle of inclination . The incline is given an acceleration a to keep the block stationary w.r.t incline plane. Then a is equal to

Taking incline plane as frame of reference, Pseudo force ma act towards right.

component of Pseudo force up the incline is m.a.cos α which is balanced by m.g.sin α. so,

m.a.cos α = m.g.sin α

⇒a= g tan α

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Question 15:

moderate

A fireman wants to slide down a rope. The rope can bear a tension of (3/4)th of the weight of the man. With what minimum acceleration should the fireman slide down:

According to the question, T=3mg/4

Using equations of motion, i.e. Fnet = m.a

⇒mg-T =ma

⇒mg-3mg/4= ma

⇒a =g/4

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Question 16:

moderate

A block of mass 10 kg is suspended through two light spring balances as shown in figure

Always remember that spring balance measure Tension in the string in form of kg-wt. As tension in the string is 10 kg-wt.

Reading of Both the scales will read 10 kg

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Question 17:

moderate

A trolley of mass 8 kg is standing on a frictionless surface inside which an object of mass 2 kg is suspended. A constant force F starts acting on the trolley as a result of which the string stood at an angle of 37⁰ from the vertical (bob at rest relative to trolley) Then:

Using concept of Pseudo force

a= g tanθ

⇒ a = g tan 37 = 10 ×3/4= 7.5 m/s^2

Using Equation of motion

⇒Fnet = 10×2.5 = 25 N

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Question 18:

moderate

Consider a car moving on a straight road with a speed of 10 ms^-1. The distance at which car can be stopped is: [μ=0.5]

Friction force action on the object is f=μ.N= μ.mg

Retardation of the block is a=μ.g = 0.5× 10 =5

Using equation of motion. Stopping distance = u^2/2a=(10 × 10)/( 2* 5)=10 m

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Question 19:

difficult

A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

Maximum Friction force action on an object is μN.

Here μ=0.2 and Normal reaction force will be equal to applied force i.e 10 N.,

So, friction force is 0.2*10 = 2N

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Question 20:

moderate

A marble block of mass 2 kg lying on ice when given a velocity of 6 ms^-1 is stopped by friction in 10 s. Then the coefficient of friction is

Using equation of motion

v= u + at

⇒ 0= 6 + a ×10

⇒ a= -0.6 m/sec^2

Maximum Friction force is μN = μ mg so, a= μ.g

⇒ -0.6 = - μ. 10

⇒μ = 0.06

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