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 700g, 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.

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  =

= [ kr²/2]₀^-4

= 1/2 kX16X10⁻⁴

= .5 x 1715 x 10² x 16 x 10⁻⁴

= 137.20 J

b

explanation:

bc its just b

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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