A 20-N crate starting at rest slides down a rough 5.0-m long ramp,inclined at 25° with the horizontal.30 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?
A) 0.98 m/s
B) 2.1 m/s
C) 3.5 m/s
D) 4.7 m/s

Question

Quizplus · Accepted Answer

The work done by gravity is: Wg = mgh = (20 kg)(9.8 m/s^2)(5.0 m) = 980 JThe work done by friction is: Wf = -30 JThe net work done on the crate is: Wnet = Wg + Wf = 980 J - 30 J = 950 JThe change in kinetic energy of the crate is: KE = 1/2 mv^2The final speed of the crate is: v = sqrt(2KE/m) = sqrt(2(950 J)/(20 kg)) = 3.5 m/s

A 20-N Crate Starting at Rest Slides Down a Rough