# Work and Kinetic Energy Problem

1. Sep 27, 2008

### maniacp08

A 6.0kg block slides 1.5m down a frictionless incline that makes an angle of 60 degrees with the horizontal.

a) Find the work done by each force when the block slides 1.5m(measured along the incline)
b) What is the total work done on the block?
c) What is the speed of the block after it has slid 1.5m, if it starts from rest?
d) What is its speed after 1.5, if it starts from initial speed of 2.0m/s?

I drew a free body diagram and there are total of 3 forces.
Normal Force = 58.86 cos 60 = 29.43N
Force of gravity = 6.0 * -9.81 = -58.86N
Force parallel = 58.86 sin 60 = 50.97N

Work = F cos Theta * displacement

Do I need to incorporate the Cos Theta in the work equation below?
Will the work for these forces be:
Normal force = 29.43N * 1.5
Force of gravity = -58.86N * 1.5
Force Parallel = 50.97N * 1.5

Will the answer for B - the total work done be the sum of the Work from all the forces?

Is the answer for C related to F = MA?
Force parallel = M A
50.97N = 3 * A
A = 16.99 m/s^2

Using Vf^2 = Vi^2 + 2ad
Vf^2 = 0 + 2(16.99)(1.5)
Vf^2 = 50.97
Vf = 7.13 m/s

And for question D, is same kinematics equations but substitute 2.0m/s for initial velocity?

Any help would be great! Thanks.

2. Sep 27, 2008

### Galileo's Ghost

Hmmmm...whenever I try to identify forces in a situation like this one, I ask myself "what is the object that is responsible for pushing or pulling the block?" Remember a component of a force isn't a separate force all by itself!

Absolutely! And remember, the "theta" in this equation is NOT the angle of the incline. It is related to the relationship between the direction of the force and the direction of displacement.

Yes

Yes. Your analysis for C and D looks good.

3. Sep 27, 2008

### maniacp08

Since there is no force acting upon the object to slide down, the only force would be the force of gravity and it would be the parallel force but the parallel force is not considered a force?

Would you explain a bit more please, Im curious now.

It would be Cos 0 in this case correct?

4. Sep 27, 2008

### LowlyPion

How much work is done by the normal force on the block?

What is Sinθ times the distance along the incline? Do you recognize what that is?

For part c. If the block benefits from work, what does that translate into? (Hint: The title of your problem.)

Part d. If your block has kinetic energy already then what does the additional work result in?

5. Sep 27, 2008

### Galileo's Ghost

Gravity IS the force causing the object to slide down. What you are calling the "parallel force" is a component of the gravitational force, it is not a seperate force unto itself. It is the portion of the gravitational force acting in the direction of the incline.

That depends. Cos 0 would apply if the force in question is in the direction of displacement. So, this would not be appropriate for something like the Normal force you identified.

6. Sep 27, 2008

### maniacp08

So the forces are the Normal Force and Gravity.

The work done by gravity would be
-58.86N * Cos 0 * 1.5?

And the work done by Normal Force would be
58.86 Cos 60 * Cos 90 * 1.5?

It translate into Kinetic Energy
F * displacement = K(final) - K(initial)
So K(final) = F * Displacement + K(initial)?

Im sorry, Im confused on those two parts.