How is work calculated on an inclined plane for a block of ice?

[gracedescent]

Homework Statement

The block of ice weighs 500N. It is being pushed up an inclined plane with a height of 3m and a hypotenuse of 6m.

1a. How much force is need to push it to the top?

1b. How much work is required to push it to the top?

Homework Equations

W=Fd=deltaKE=deltaGPE

The Attempt at a Solution

I have no idea for 1b. because W=Fd, the force is 500N, and d would have to equal 3m to get the 1,500J needed. However, if this is true, and the force is parallel to the motion, why wouldn't you use the hypotenuse instead of the height for the distance?

1b. GPE at the top = (500N)(3m) = 1,500J
Thus, W = Fd
1,500 J = F(6m) = 250N
Is this correct?

Thanks!

 

[PhanthomJay]

Homework Statement

The block of ice weighs 500N. It is being pushed up an inclined plane with a height of 3m and a hypotenuse of 6m.

1a. How much force is need to push it to the top?

1b. How much work is required to push it to the top?

Homework Equations

W=Fd=deltaKE=deltaGPE

The Attempt at a Solution

I have no idea for 1b. because W=Fd, the force is 500N, and d would have to equal 3m to get the 1,500J needed. However, if this is true, and the force is parallel to the motion, why wouldn't you use the hypotenuse instead of the height for the distance?

1b. GPE at the top = (500N)(3m) = 1,500J
Thus, W = Fd
1,500 J = F(6m) = 250N
Is this correct?

Thanks!

You are arriving at the corect answer, but I'm not sure if you understand why. First, your relevant equation is incorrect. Look up the work energy theorem and the conservation of total energy principle. Conservation of energy usually works best for finding work when gravity and applied forces (or friction forces) are at play.