Box Sliding Down Incline

Question

A 2.0 Kg box is sliding down a 20 degree, 5-m long ramp with coefficient of kinetic friction 0.2. if it started at 1.0 m/s, how fast will it be at the bottom of the ramp?

Is there anyway I can get this solved out completely ? I have an answer but I have no idea if it is right or not. Like If you can plug in the numbers.

Answer 1

There are two ways of solving it -- from Newton's Second Law and from

conservation of energy. Both give the same answer.

I think the latter is a little easier here, so I will use that.

Let

m = 2.0 kg be the mass of the box,

α = 20° the be incline of the ramp,

L = 5 m be the length of the ramp,

f = 0.2 be the coefficient of kinetic friction,

v₀ = 1 m/s be initial velocity,

v₁ (to be computed) be the final velocity,

g = 9.81 m/s² be gravitational acceleration.

The height of the ramp is Lsin(α).

The weight of the box is mg.

The component of the weight normal to the ramp is mgcos(α).

The force of friction if fmgcos(α).

By conservation of energy,

the initial potential energy -- mgLsin(α), plus

the initial kinetic energy -- mv₀²/2, must be equal to

the final kinetic energy -- mv₁²/2, plus

the work friction done during the slide -- Lfmgcos(α).

That is,

mgLsin(α) + mv₀²/2 = mv₁²/2 + Lfmgcos(α)

v₁ = √(2Lg(sin(α) - fcos(α)) + v₀²)

Substituting actual numbers

v1 = √(2×5×9.81×(sin(20°) - 0.2cos(20°)) + 1²) ≈ 4 m/s

Please note that the result is independent of m.