Question

A​ heavy-duty shock absorber is compressed 4 cm from its equilibrium position by a mass of 700nbspkg. How much work is required to compress the shock absorber 4 cm from its equilibrium​ position? (A mass of 700 kg exerts a force​ (in newtons) of 700​g, where galmost equals9.8 m divided by s squared​.)Set up the integral that should be used to find the work required to compress the shock absorber 4 cm from its equilibrium position. Use decreasing limits of integration. Express all displacements in meters.

Answer 1

Answer:

Explanation:

A mass of 700 kg will exert a force of

700 x 9.8

= 6860 N.

Amount of compression x = 4 cm

= 4 x 10⁻² m

Force constant K = force of compression / compression

= 6860 / 4 x 10⁻²

= 1715 x 10² Nm⁻¹.

Let us take compression of r at any moment

Restoring force by spring

= k r

Force required to compress = kr

Let it is compressed  by small length dr during which force will remain constant.

Work done

dW =  Force x displacement

= -kr -dr

= kr dr

Work done to compress by length d

for it r ranges from 0 to -d

Integrating on both sides

W  = $\int\limits^{-4}_0 {kr} \, dr$

= [ kr²/2]₀^-4

= 1/2 kX16X10⁻⁴

= .5 x 1715 x 10² x 16 x 10⁻⁴

= 137.20 J