A block of mass m = 4 kg is dragged 2 m along a horizontal surface by a force F = 30 N acting at `53^(@)` to the horizontal. The initial speed is `3 m//s and mu_(k) = 1//8`.
Find its final speed

Explanation

(a) The force acting on the block are shown in the figure. Clearly,` W_(N) = 0 `and `W_(g) = 0`, where,
`W_(F) = Fs cos theta`
`W_(f) = - fs = - mu_(k)Ns`
where N = `mg - F sin theta`
By work energy theorem, change in kinetic energy is the algebraic sum of work done by all forces.
`Delta K = Fs cos theta - mu _(k)(mg - F sin theta)s`
[Since `N = (mg - F sin theta)] = 30(2)(0.6)-1/8(40-24)(2) = 32 J`
(b) Now `DeltaK = 1/2 mv_(f)^(2) - 1/2 mv_(1)^(2) = 32 J`
Putting `v_(i) = 3 m/s, we get v_(f) = 5 m/s`