Question

18 Consider the system shown in the figure.
Coefficient of friction between the block and table
is u = 0.5. The system is released from rest. Find
the work done by friction, when the speed of block
is 10 m/s (m = 1 kg)
(1) -10 J
(2) -20 J
(3) -30 J
2m
(4) -50 J

Answer 1

Here is the solution and figure

Answer 2

The work done by friction is - -50 J.

Let the displacement of the block be  "s". Then,

workforce = −fs .s

2Mg−T=2mg ____ (i)

T−μmg=ma

⇒T−   mg /2 =ma ___ (ii)

Adding (i) and (ii), we get

3ma=  3mg/2

​ a=g/2

To find displacement at v = 10 m/s

Applying 3rd  equation  of motion,

v^2 −u^2   =2as

⇒(100)−0=2×  2 g s

⇒s=  100/g  =  100 /10

​ =10m

work-done by friction = (−  mg/2  )s

= − 10/2 ×10

= −50J

Thus, the correct answer is option 4.  -50 J