Question

A bloke of mass 5 kg is dropped from a height 5 m over a board which is supported by two parallel and same
spring constant 100 N/m.find the work done by each sprig
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Answer 1

Answer:125 J

Explanatiion:

2)In order to find the Kinetic enrgy of block we apply the conservation of energy here

Potential energy from a height of 5m=Kinetic energy on reaching springs

mgh=$\frac{1}{2}$m$v^{2}$       --------(2)

W=ΔKE

W =1/2m$v^{2}$-0

W=mgh            [We know this from (2)}

W=5 x g x  5m      [g=10m/s^2]

W=250 J

According to work energy theorem

1)Work done by force on moving object=Change in its Kinetic energy

W=ΔK.E.

The work done by springs is equal to the change in kinetic energy of the falling block.

here change in KE is 0-

0 is the final KE since the block is coming to rest once it hits and compresses  the spring board and all that kinetic energy is stored as the springs potential energy.

Since each spring in parallel gets compressed by same extent

we can say that both equal amount of work

That is total work done=2 x Work done by 1 spring

Therefore 250=2 x required answer

required answer=250/2= 125 J