Question

A 0.019 kg block on a horizontal frictionless surface is attached to a string whose spring/force/elastic constant k is 120 N/m. The block is pulled from its equilibrium position at x=0 m to a displacement x=+0.080 m and is released from rest. The block then executes simple harmonic motion along x-axis (horizontal). When the displacement is x=0.051 m, what is the kinetic energy of the block in J?

Answer 1

Mass of block m= 0.019kg

Spring constant k = 120N/m

Maximum compression A= 0.08m

Potential energy in spring of x distance u = kx²/2

Now using energy conservation

Total energy of block when it is compressed to

A = 0.08m

Potential energy of spring u = kA²/2

Kinetic energy k = 0 as velocity at A(maximum compression) is zero.

So E_total = kA²/2

Now energy at x= 0.051

U = kx²/2

Let kinetic energy at this point is w

E_total = kx²/2 + w

As energy is conserved E_total is constant.

kx²/2 + w = kA²/2

120×(0.051)²/2 + w = 120×(0.08) ²/2

0.156 + w = 0.384

W= 0.228J