Question

Pleaed do it ASAP

06. A block of mass 2.0 kg is dropped from a height h onto a spring of spring constant 1960 N/m. The compression of the spring is 10 cm. a. Find h. b. Then the block is further pushed down an additional 10 cm and released. What is the elastic potential energy of the compressed spring just before the release?

Answer 1

Given : mass of block ,m=2 kg

Spring constant,k=1960N/m

Compression,x=10cm=0.1 m

Solution: Here , potential energy of block before falling is transferred to spring and using this Energy,spring compress.

a)Hence, initial Potential energy = Potential energy due to spring compression (U)

Initial P.E=mg(h+x) and U=(1/2)×K×x²

mg(h+x)=0.5×k×x²

2×9.8(h+0.1)=0.5×1960×0.01

h+0.1=98÷(19.6)=5

h=5-0.1=4.9m

b)When block is pushed further 10cm then total compression in the spring is x' , (10+10)cm=20cm=0.2m

Therefore,Elastic potential energy before release,U=1/2×K(x')²

x'=20cm=0.2m and k=1960N/m

So,U=(1/2)×1960×0.2×0.2=39.2J