please show work

Question

A 40.0-N crate starting at rest slides down a rough 6.00-m-long ramp, inclined at 30.0∘ with the horizontal. The magnitude of the force of friction between the crate and the ramp is 6.0 N. What is the speed of the crate at the bottom of the incline?

(a) 1.60 m/s

(b) 3.32 m/s

(c) 4.5 m/s

(d) 6.42 m/s

 (e) 7.75 m/s

Answer 1

Let

g = 9.81 m/s² be gravitational acceleration,

m = 40/9.81 kg be the mass of the block,

s = 6 m be the length of the ramp,

h = 6sin(30°) = 3 m be the height of the ramp,

F = 6 N be the force of friction,

v (to be calculated) be the speed at the bottom.

The potential energy at the top (mgh) gets converted

into the kinetic energy at the bottom (mv²/2)

plus the work of friction (Fs).

mgh = mv²/2 + Fs

v = √(2(mgh - Fs)/m)

= √(2(40×3 - 6×6)/(40/9.81)) = 6.42 m/s