A block of mass 2 kg, initially at rest on a horizontal floor, is now acted upon by a horizontal force of `10 N`. The coefficient of friction between the block and the floor is 0.2. If `g = 10 m//s^(2)`, then :

Options

A. The work done by the applied force in 4 s is 240 J.
B. The work done by the firctional force in 4s is 0,
C. The work done by the net forces in 4s is 0.s
D. The change in kinetic energy of the block in 4s is 144 J.

Answer

A, D

Explanation

`f_(L) = mu N = 0.2 xx 2 xx 10 = 4N`
`a = (F_("net"))/(m) = (10 - 4)/(2) = 3 m//s^(2)`
`s_(4s) = (1)/(2) xx 3 (4)^(2)= 24 m`
`W_("app. Force") = 10 xx 24 = 240 J`
`W_(f) = -4 xx 24 = - 96 J`
`W_("net") = DeltaK = 240 - 96 = 144 J`