A 25 kg box slides, from rest, down a 9-m-long incline making an angle 15 degrees with the horizontal. The speed of the box when it reaches the bottom of the incline is 2.40 m/s.

Question

(a) What is the coefficient of kinetic friction between the box and the surface of the incline?
(b) How much work is done on the box by the force of friction?
(c) What is the change in the potential energy of the box?

Answer 1

(a)
Distance, S = 9m
initial velocity, u = 0 m s^-2
final velocity, v = 2.4 m s^-2
acceleration, a (assumed uniform)
= (v²-u²)/2S
= (2.4²-0²)/(2*9)
= 0.32 ms^-2

Acceleration due to gravity, g
= 9.8 m s^-2
component of g along incline
= g sin(θ)
= 9.8 sin(15°)
= 2.536 m s^-2
Therefore reduced acceleration due to friction, af
= 2.536-0.32
= 2.216 ms^-2

Frictional force, F
= m(af)
= 25 kg * 2.216 m s^-2
= 55.411 N
Normal force acting on inclined plane, N
= m * g cos(θ)
= 25 * 9.8 * cos(15°)
= 236.6 N
Coefficient of kinetic friction, μ
= F/N = 0.234

(b) work done = force * distance
(c) Change in potential energy = mgh
where h is the difference in elevation (final - initial).

Answer 2

if the mass were suspended by water and it's specific gravity were less than one would it knot float? slip up!