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Friction Force NAT Level - 1 - Physics MCQ


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10 Questions MCQ Test - Friction Force NAT Level - 1

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*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 1

A block of mass 4kg is kept on ground. The coefficient of friction between the block and the ground is 0.80. An external force of magnitude 30N is applied parallel to the ground. The resultant force (in N) exerted by the ground on the block is : (=10 m/s2)


Detailed Solution for Friction Force NAT Level - 1 - Question 1


The correct answer is: 50

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 2

A block of mass 15kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50N. The coefficient of friction between the surface of contact is  (= 10 m/s2)


Detailed Solution for Friction Force NAT Level - 1 - Question 2

The free body diagram of the block is
N is the normal reaction exerted by inclined plane on the block.
Applying Newton’s second law to the block along the normal to the incline.



Solving we get μ = 1/2
The correct answer is: 0.5

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*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 3

An external force of 6 N is applied on body at rest at time t = 0.  At  t = 1 sec direction of force is reversed. Find the speed ms–1 of body when it comes back to the initial position.


Detailed Solution for Friction Force NAT Level - 1 - Question 3



The correct answer is: 1.095

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 4

Figure shows a block kept on a rough inclined plane. The maximum external force along the plane downwards for which the block remains at rest is 1N while the maximum external force along the incline upwards for which the block is at rest is 7N. The coefficient of static friction μ is :


Detailed Solution for Friction Force NAT Level - 1 - Question 4

When block is about to slide down

When block in about to slide up

From (1) and (2) 

The correct answer is: 0.769

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 5

What is the maximum value of the force F (in N) such that the block shown in the arrangement does not move?


Detailed Solution for Friction Force NAT Level - 1 - Question 5





The correct answer is: 20

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 6

A block is given velocity 10 m/s along the fixed inclined as shown in figure from the bottom of the inclined plane. Find the total distance traveled (in meter) along the inclined plane, if the coefficient of friction is equal to 0.5 :


Detailed Solution for Friction Force NAT Level - 1 - Question 6

distance traveled along inclined plane till it stops

But friction force is not sufficient to stop there, so it will come back to the bottom of the inclined plane.
Total distance traveled  5 + 5 = 10m

The correct answer is: 10

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 7

A uniform rope so lies on a table that part of it lays over. The rope begins to slide when the length of  hanging part is 25% of entire length. The coefficient of friction between rope and table is :


Detailed Solution for Friction Force NAT Level - 1 - Question 7



The correct answer is: 0.33

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 8

With reference to the figure shown, if the coefficient of friction at the surface is 0.42, then the forces F (in N) required to pull out the 6.0kg block with an acceleration of 1.50 m/s2 will be :


Detailed Solution for Friction Force NAT Level - 1 - Question 8


The correct answer is: 51

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 9

Starting from rest a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is :


Detailed Solution for Friction Force NAT Level - 1 - Question 9


Let acceleration in 1st case is a1 and that in second case is a2



From (1), (2) and (3), we get
μ = 0.75
The correct answer is: 0.75

*Answer can only contain numeric values
Friction Force NAT Level - 1 - Question 10

A body of mass 10kg lies on a rough horizontal surface. When a horizontal force of F newtons acts on it, it gets an acceleration of 5 m/s2. When the horizontal force is doubled, it gets an acceleration of 18 m/s2. The coefficient of friction between he body and the horizontal surface is :


Detailed Solution for Friction Force NAT Level - 1 - Question 10


From (1) and (2)
μ = 0.8
The correct answer is: 0.8

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