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# A constant force does 12.0 Joules of work pushing a 350 gram block a distance of 2.5 meters up a ramp. The blo?

A constant force does 12.0 Joules of work pushing a 350 gram block a distance of 2.5 meters up a ramp. The block starts from rest and reaches a speed of 4.75 m/s after being pushed for 2.5 m. The block’s potential energy increases by 5.0 J as it travels this distance. (a) What is the net work (also called the total work) done on the block? (b) What is the coefficient of friction between the block and the ramp?

### 1 Answer

- jcherry_99Lv 79 years agoFavorite Answer
The key is to start with the PE. If the PE increases by 5J it means that the vertical height can be calculated.

PE = 5 J

m = 350 grams * [1 kg / 1000 grams] = 0.350 kg

g = 9.81 m/s^2

h = ??

h = Pe / (g * m)

h = 5 / (9.81 * 0.35)

h = 1.45 meters.

The Kinetic Energy is

KE = 1/2 m*v^2

Ke = 1/2 * 0.35 * 4.75^2

Ke = 3.95 J

The work done on the block is 5 + 3.95 = 8.95 J

Since 12 J was needed to push the block up the ramp 3.05 J [ 12 - 8.95] of work was done against friction.

The normal force on the block = mg*sin(theta)

sin(theta) = height / hypotenuse = 1.45 meters/ 2.5 meters = 0.58

m = 0.35

g = 9.81

sin(theta) = 0.58

Normal = 1.99 N

Work done by frictional force = 3.05 J

Frictional force = ???

height = 1.45 m

W = F*h

3.05 = F *1.45

F = 2.01 N

Finally

F = u*N

2.01 = u*1.99

u = 1.0

which is impossible but I invite you to check my work.