A particle of mass M moves along a horizontal x axis from x = 0 to x = L. The coefficient of kinetic friction varies as a function of x as k(x) = 0 - x, where 0, are constants of appropriate dimensions, so that k(L) = 0. The total work done by the frictional force during the motion is n 0MgL, where g is the acceleration due to gravity. The value of n is:
- 12
- 3
- 1
- 13
Answer
(1) 12
Solution
Given k(L) = 0 0 - L = 0 = 0L.
The magnitude of work done by friction is:
W = 0L k(x)Mg dx = Mg 0L ( 0 - 0Lx )dx = Mg [ 0 L - 0 L22L ] = 12 0 MgL.
Comparing with n 0 MgL, we get n = 12.
Answer: (1)