A 10.0 kg block of wood is shot up an inclined ramp

Question

A 10.0 kg block of wood is shot up an inclined ramp (from the bottom) by a spring which gives the block an initial velocity of 6.40 m/s up the ramp. The angle of the ramp is 30o. The coefficient of kinetic friction is 0.560

a) Determine the maximum distance up the ramp that the block travels.

b) Determine if the crate will slide back down the ramp. (Include your reasoning).

Answer 1

Let

m = 10 kg be the mass of the block,

v = 6.4 m/s be the initial velocity,

α = 30° be the angle of the ramp,

f = 0.56 be the coefficient of kinetic friction,

g = 9.81 m/s² be gravitational acceleration.

The force of gravity is of magnitude mg, and is downward.

It can be considered as the sum of two forces --

B = mgsin(α) is the force parallel to the ramp pulling the block back,

N = mgcos(α) is the force perpendicular to the ramp, causing friction.

The force of friction is

F = Nf = mgcos(α)f

and acts always against the direction of movement.

a)

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

Conservation of Energy.

I will show you both.

Newton's Second Law:

Both forces B and F act to slow the block down. So the block is under deceleration

a = (B+F)/m = g(sin(α) + cos(α)f)

It will stop in time

t = v/a = v/g(sin(α) + cos(α)f)

In that time it will travel distance

s = at²/2

= g(sin(α) + cos(α)f)v²/2g²(sin(α) + cos(α)f)²

= v²/2g(sin(α) + cos(α)f)

Conservation of Energy:

At the outset the block has kinetic energy

mv²/2

It will stop when that kinetic energy gets converted to its potential energy plus

the work of friction.

The potential energy after travelling distance s is

mgsin(α)s

The work of friction after travelling distance s is

Fs = mgcos(α)fs

Conservation of Energy is expressed by

mv²/2 = mgsin(α)s + mgcos(α)fs

s = v²/2g(sin(α) + cos(α)f)

Substituting actual numbers

s = 6.4²/(2×9.81×(sin(30°) + cos(30°)×0.56)) ≈ 2.1 m

b)

The question cannot be answered without knowing the coefficient of static friction.

So I will just calculate whether the block is capable of sliding back,

once it started.

On the way down the force B is the same as in case a),

but the force of friction F now acts against B.

So the box will slide down, provided gravity, represented by B,

is stronger than friction, represented by F.

That is

B > F

mgsin(α) > mgcos(α)f

sin(α) > cos(α)f

Substituting actual numbers

sin(30°) > cos(30°)×0.56

0.5 > 0.48

The conclusion is that the box will slide back, provided static friction

does not causes it to remain at the highest point.