Q 1 :    

A 2 kg brick begins to slide over a surface which is inclined at an angle of 45° with respect to horizontal axis.

The co-efficient of static friction between their surfaces is         [2024]

  • 1

     

  • 0.5

     

  • 1.7

     

  • 13

     

(A)  Slipping starts when, tanθ=μs (θ is angle of respose) 

      μs=tan 45o

      μs=1

 



Q 2 :    

Given below are two statements:

 

Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.

 

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.

 

In the light of the above statements, choose the most appropriate answer from the given below:            [2024]

  • Statement I is correct but Statement II is incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are incorrect

     

  • Both Statement I and Statement Il are correct

     

(B)  Co-efficient of friction depends on surface in contact So, it depends on material of object.

 



Q 3 :    

A heavy box of mass 50 kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is 0.3, then the force of kinetic friction is          [2024]

  • 1.47 N

     

  • 14.7 N

     

  • 1470 N

     

  • 147 N

     

(4)

Given m=50 kg, μk=0.3        

fk=μkN=μkmg=0.3×50×9.8=147 N

 



Q 4 :    

A given object takes n times the time to slide down 45° rough inclined plane as it takes the time to slide down an identical perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is ______            [2024].

  • 1-n2

     

  • 1-1n2

     

  • 1-n2

     

  • 1-1n2

     

(4)

Case-1: No friction

a=gsinθ

=12(gsinθ)t12t1=2gsinθ

Case-2: With friction

a=gsinθ-μgcosθ

=12(gsinθ-μgcosθ)t22t2=2(gsinθ-μgcosθ)

t2=nt12(gsinθ-μgcosθ)=n2gsinθ

(sinθ-μcosθ)=1n2sinθ

(1-μ)=1n2μ=1-1n2



Q 5 :    

A block of mass m is placed on a surface having a vertical cross-section given by y=x2/4. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is     [2024].

  • 1/4 m

     

  • 1/2 m

     

  • 1/6 m

     

  • 1/3 m

     

(1)

fs=mgsinθ                    ...(1)

N=mgcosθ                  ...(2)

Now, fsμN

mgsinθμmgcosθ

tanθμ                         ...(3)

According to question, y=x24

tanθ=dydx=14·2x=x2

Put in (3) x2μx2(0.5)

x1 So y124ymax=14=0.25



Q 6 :    

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure.

If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent the block from sliding down, then the value of |F1|-|F2| is        [Use g = 10 m/s2]                    [2024]

  • 253 N

     

  • 53 N

     

  • 532 N

     

  • 10 N

     

(2)

fK=μmgcosθ

=0.1×50×32=2.53 N   

F1=mgsinθ+fK=25+2.53

F2=mgsinθ-fK=25-2.53

 F1-F2=53 N



Q 7 :    

Consider a block and trolley system as shown in the figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in ms-2 is ______.

(Consider that the string is massless and unstretchable and the pulley is also massless and frictionless.)           [2024]

  • 3

     

  • 4

     

  • 2

     

  • 1.2

     

(3)

fk=μN=0.04×20g=8 Newton

Acceleration a=60-826=2 m/s2