A 5.0kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0m down the plane. Assuming there is no friction between the block and the surface, calculate:

1. The gravitational potential energy at the top of the plane,

2. The component of the weight parallel to the plane,

3. The acceleration of the block,

4. The velocity of the block at the bottom of the plane,

5. The kinetic energy at the bottom of the plane.

Since there's no friction, the change in potential energy will be equal to the kinetic energy at the end. Am I right in thinking that the work done on the block is also equal to the kinetic energy gained by the block?

If so, that gives:

$$W = E_k = \Delta E_p$$

$$F \cdot s \cdot \cos\theta = \frac{m \cdot v^2}{2} = m \cdot g \cdot \Delta h$$

Substituting in values gives:

$$98.1 \cos \theta = 2.5v^2 = 49.05 \Delta h$$

Without knowing the angle the slope is on, I'm not sure where to go from here.

I've been assuming that the block travels at an angle, 2 metres down the slope. Perhaps it goes down 2 metres vertically?

$$49.05 \cdot s \cdot \cos \theta = 2.5v^2 = 98.1$$

This would help with parts 1, 4, and 5; but I wouldn't know how to answer parts 2 and 3.

The wording of the question implies that the 2.0m is measured along the incline. However the angle of inclination $\theta$ is not given so it is not possible to calculate the loss in gravitational PE.

Without a value for $\theta$ you cannot answer parts 2 and 3 either. None of the 5 parts can be answered without $\theta$. The question is faulty. Essential information is missing.

edited Sep 7 by sammy gerbil