Q 1 :

The work done on a particle of mass m by a force, K[x(x2+y2)3/2i^+y(x2+y2)3/2j^]  (K being a constant of appropriate dimensions), when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the xy plane is               [2013]

  • 2Kπa

     

  • Kπa

     

  • Kπ2a

     

  • 0

     

(4)

Radius of circular path =a

The equation of circle is x2+y2=a2

Given : force

F=K[xi^(x2+y2)3/2+yj^(x2+y2)3/2]

F=K[xi^(a2)3/2+yj^(a2)3/2]

[IMAGE 106]

F=Ka3[xi^+yj^]

The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path.

Here the angle between the force which acts radially outwards and displacement which is tangential to the circular path is 90°

 Work done, W=FScosθ=0



Q 2 :

If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3.                      [2003]

[IMAGE 107]

  • W1>W2>W3

     

  • W1=W2=W3

     

  • W1<W2<W3

     

  • W2>W1>W3

     

(2)

In a conservative field work done does not depend on the path i.e., path independent. The gravitational field is a conservative field.

 W1=W2=W3



Q 3 :

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g=10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.                   [2009]

[IMAGE 108]



(8)

[IMAGE 109]

When the system is released,

T-mg=ma    ...(i)

Mg-T=Ma    ...(ii)

From eq. (i) & (ii)

a=(M-m)gM+m=g3

and T=4mg3

For block m=0.36 kg

u=0,  a=g3,  t=1,  s=?

s=ut+12at2=0+12×g3×12=g6  (m=0.72 kg)

 Work done by the string on m

W=Tscos0°=4mg3×g6×1=4×0.36×10×103×6=8J



Q 4 :

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force F=(αyi^+2αxj^) N, where x and y are in meter and α=-1 N m-1. The work done on the particle by this force F will be _____ Joule.                  [2019]

[IMAGE 110]



(0.75)

Given : Force, F=(αyi^+2αxj^)

and α=-1 N m-1

We know that dW=F·dr=(αyi^+2αxj^)·(dxi^+dyj^)

 dW=αydx+2αxdy

Work done

From AB, dy=0, as y=1

 W1=01αydx=α01dx=α

From BC, dx=0, as x=1

 W2=10.52αxdy=10.52αdy=2α(0.5-1)=-α

From CD, dy=0, as y=0.5

 W3=10.5α×0.5dx=α2(0.5-1)=-α4

From DE, dx=0, as x=0.5

 W4=0.502α×0.5dy=α(0-0.5)=-α2

From EF, dy=0, as y=0

 W5=α×0dx=0

From FA, dx=0, as x=0

 W6=2α×0dy=0

 Total work done W=W1+W2+W3+W4+W5+W6

W=α-α-α4-α2=-3α4=-3(-1)4=0.75J