First slide

Work

Question

A spring of force constant $150 {Nm}^{- 1}$ is stretched through 10cm. The energy stored in the spring is

Moderate

A

1.50J
B

1500J
C

750J
D

0.75J

Solution

$\begin{array}{l} U = \frac{1}{2} K x^{2} = \frac{1}{2} \times 150 \times {(10^{- 1})}^{2} = 75 \times 10^{- 2} \\ U = 0.75 J \end{array}$

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