You get a running start, pushing a box of mass 21 kg up to a ramp of grade 0.7 and friction coefficient 0.48. The box has a speed 10.9 m/s when it starts up the ramp, the same time that you let it go. The ramp ends at a platform of height 3.14 m.(a) What is the work done by friction as the box goes up the ramp?(b) What is the speed of the box just before it hits the ground?

a) -442.6 J

b) 8.8 m/s

Explanation:

a)

The work done by the friction force is equal to:

where

is the frictional force on the box, where

is the coefficient of friction

m = 21 kg is the mass of the box

is the acceleration due to gravity

is the angle of the ramp

The negative sign is due to the fact that the frictional force is opposite to the motion of the box

d is the length of the ramp

So the work can be rewritten as

Here we know that the grade of the ramp, which is the ratio between height and horizontal length, is 0.7:

This means that

So we find the angle:

We also know that the height of the ramp is

h =  3.14 m

So we can find the length of the ramp:

Therefore, the work done by friction is

b)

The final speed of the box can be found by using the law of conservation of energy. In fact, the initial kinetic energy of the box (at the bottom) + the work done by friction must be equal to the final kinetic energy of the box, as it reaches  the ground after leaving the platform.

So we can write:

where:

u = 10.9 m/s is the initial speed of the box

m = 21 kg is the mass of the box

is the work done by friction

is the final speed of the box as it reaches the ground

Solving for v, we find:

the part of the wave he is travelling on is the crest

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west