Question

A lead pallet of mass 10g leaves an air gun with a velocity of 40 ms^-1.what is the magnitude of potential energy stored by its spring?

Answer 1

Step 1: Given data,

1. Mass of pallet =$10g=\frac{10}{1000}=\frac{1}{100}kg$
2. The velocity of the pallet = 40 ms^-1

Step 2: Calculate the potential energy of the spring,

The potential energy stored in the spring is converted into the kinetic energy of the lead pallet.

KE of pallet = PE of spring

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times \frac{1}{100}\times {\left(40\right)}^2 \\ =8J \end{gathered}$$

$\because$KE of pallet = PE of spring

Thus, the potential energy stored in the spring is PE = 8 J.