Question

15. a 12 kg bowing ball dropped
from height of 10 m onto a
spring. If the spring is to
compressed 2 m before halting
the ball, whar is the spring
constant k of the spring?​

Answer 1

Answer : 600 N-m

Potential energy (U.E) = MgH

Potential energy (U.E) = MgHSpring Potential Energy : 1/2.k.x²

According to question,

$→MgH = \frac{1}{2} k. {x}^{2} \\ \\ →12 \times 10 \times 10 = \frac{1}{2} .k. {2}^{2}$

$→k = 600 \: Nm$

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Explanation:

As the ball descends towards the spring,It's Gravitational Potential Energy gets converted into Kinetic Energy.

When the ball finnally touches the spring, it's kinetic energy starts being stored into the spring in the form of Spring Potential Energy.

Therefore, if we neglect any loss of energy in between,

we can say that,

The gravitational potential energy due to its height gets converted to Spring Potential Energy in the end.