Q 1 :    

A force (3x2+2x-5)N displaces a body from x=2m to x=4m. Work done by this force is __________ J.             [2024]



(58)       Force, F=(3x2+2x-5)N,x1=2m and x2=4m

              Work done, W=x1x2F·dx=24(3x2+2x-5)dx

               W=[x3]24+[x2]24-5[x]24=[64-8]+[16-4]-5[4-2]=58J

 



Q 2 :    

A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be         [2024]

  • 2 : 1

     

  • 1 : 2

     

  • 3 : 2

     

  • 1 : 1

     

(4)

Work done by gravity is path independent, it depends only on vertical displacement, so same work is done in both paths.

 



Q 3 :    

A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is    [2024]

  • 1 m

     

  • 6 m

     

  • 5 m

     

  • 2 m

     

(4)

sin30°=h10h=5

At maximum compression

using Work-Energy Theorem,

Wgravity+Wfriction+Wspring=ΔKE

mgh-μmg(2+x)-12kx2=0-0

5×10×5-12×5×10(2+x)-12100×x2=0

250-50x2-[50+25x]=0

200-50x2-25x=0

2x2+x-8=0

x=-1±12+4×2×82×2x=-1±6541.77m

Nearest answer is 2m.



Q 4 :    

A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the coefficient of friction between the surfaces is 0.4, then the work done against friction (in J) is       [2024]

  • 4200

     

  • 3900

     

  • 4000

     

  • 4500

     

(3)

Given m=100kg, μ=0.4

Friction force, f=μmg=0.4×100×10=400N

Now W=f·s=400×10=4000J

 



Q 5 :    

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60 by a force of 10 N parallel to the inclined surface as shown in the figure. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is [g=10m/s2]     [2024]

  • 53 J

     

  • 5 J

     

  • 5×103 J

     

  • 10 J

     

(2)

fk=μkN=μkmgcos60°

fk=0.1×10×12=0.5 N

Wfriction=-fk·S

                 =-(0.5)×10

Wfriction=-5 J