How Does Friction Affect Work and Energy on an Inclined Plane?

[nikhi]

Homework Statement

1) According to an object's distance vs. force (parallel) graph, what is the work done in this process as the object moves from 2 m to 8 m?
2) a book is pushed up an 8 degree inclined plane with an initial speed of 11 m/s. If the coefficient of kinetic friction between the book and the plane is 0.15, what is the distance up the incline the book will slide?

Homework Equations

W=Force*distance
friction=mu(Normal Force)
conservation of mechanical energy equation:
E1=E2 + Wf

The Attempt at a Solution

the first ones answer is 1650 J but i have no idea why!
second one: (IS WRONG, SO PLZ HELP)
.5 mv^2 = friction*distance
distance=.5v^2/ .15 *9.8 *cos 8
d= 41.5

 

[JeremyG]

the first ones answer is 1650 J but i have no idea why!

Have you attempted this question yet? Show us your working.
Perhaps show us the graph as well so we know whether you are doing it right.

.5 mv^2 = friction*distance

Kinetic energy is not only used to do work against friction. There should be a third term in this equation

 

[nikhi]

There's no graph! It was a question my teacher asked. NO one has a clue where to start!
AH i figured it out, i assumed there would be no potential energy because no height/distance was given at all, but i realize now:

.5v^2 -gy /mu * g

You solve for y:
mgsintheta + friction =ma
a= 2.82 m/s^2

v^2 = v(initial)^2 + 2ax

x is 21.4 m

correct, and thanks!

 

[gneill]

Homework Statement

1) According to an object's distance vs. force (parallel) graph, what is the work done in this process as the object moves from 2 m to 8 m?
2) a book is pushed up an 8 degree inclined plane with an initial speed of 11 m/s. If the coefficient of kinetic friction between the book and the plane is 0.15, what is the distance up the incline the book will slide?

Homework Equations

W=Force*distance
friction=mu(Normal Force)
conservation of mechanical energy equation:
E1=E2 + Wf

The Attempt at a Solution

the first ones answer is 1650 J but i have no idea why!
second one: (IS WRONG, SO PLZ HELP)
.5 mv^2 = friction*distance
distance=.5v^2/ .15 *9.8 *cos 8
d= 41.5

Hi nikhi, Welcome to Physics Forums.

I future please submit each separate question to a separate post. This avoids having overlapping discussions of different problems in a single thread.

For your first problem I don't see the distance versus force graph mentioned. It sounds like it will be required to solve the problem.

For the second problem you've missed a source of acceleration: Friction is acting, which you have accounted for, but so is a fraction of the gravitational acceleration acting in the down-slope direction. Draw the Free Body Diagram.