# Conceptual question regarding work

Here is a problem I am having some trouble with:

The solution is below, my question is, why is conservation of energy valid in this situation.

From my understanding, just before the box hits the springs, energy is conserved since its just gravity. BUT, when the box hits the springs, THERE IS A NORMAL FORCE.

1. Why don't they take the normal force into account? Its in opposite direction of the motion.

2. The force of gravity and springs are taken into account as potential energy ,am I right here?

Solution:

Short Answer: The contact force (normal force if you like) between the pan and the box is 0 because the pan has negligible mass.

Long Answer: The key point in this problem is that the pan has neglible mass. Suppose for a second that the pan had some mass $m_p$. After falling a distance of $0.5 \, m$ the box would have velocity $v_b = \sqrt{2gh}$. In the instant of the collision, momentum would be conserved. Therefore, the velocity of the block and pan $v_{bp}$ would have to follow.

$$(m+m_p)v_{bp} = mv_b \Rightarrow v_{bp} = \frac{mv_b}{m+m_p}$$

As you can see, if the pan has neglible mass ($m_p \approx 0$), the velocity of the block-pan unit is the same as before the collision. Therefore, the total energy has not been reduced and the normal force is 0.

If the pan had had mass, then yes. The normal force would have been nonzero and would have made mechanical energy not be conserved.

And yes, both gravitational and elastic (springs) potential energies are taken into account in the term "potential energy"

• I'm suppose to avoid "thank you" comments. But this reply was very thorough and I appreciate it a lot. Thanks. – Edward Newgate May 28 '15 at 1:39

$$F = ma$$