A 32 lb cinder block is dropped on a nonfunctional printer. The block falls 4 feet before hitting. How much force does the block hit with?

It is really hard to say as the printer is not a uniform density item nor uniformly stiff. If the cinder block hits a very rigid part of the printer, forces could be very high, If the block hits a softer or springier portion of the printer, the forces will be much lower.

Let's assume, for the sake of example, that the block hits the printer in a rigid spot and the printer deflects by one inch or 1/12 (0.08333...) of a foot. Let's also assume that the printer acts as an ideal spring. This is a large stretch of an assumption as we would expect the plastics of the printer case to possibly shatter or at least take some permanent deformations due to the strike.

Let's also assume that the forces are not large enough to break any part of the cinder block.

32 lb is about one slug, the unit of mass in the US common standard.

The change in gravity potential energy will equal the change in spring potential energy.

PS = PE

½kx² = mgh

the height of drop h will be four feet and one inch

the spring compression is one inch

the mass is one slug

the gravity is 32.2 ft/s²

k is the spring constant

k = 2mgh/x²

k = 2(1.0)(32.2)(4.08) / 0.08²

k = 37870 lbf / ft

as the spring is compressed only one inch, the force seen by the block is

F = kx

F = 37870(0.083)

F = 3155 lbf

So we can see that the force applied to the stiff portion of the printer is nearly 100 times the actual rest weight of the cinder block.

If we have the block strike a softer part of the printer, say it deflects four inches before the cinder block stops, the results would be

k = 2(1.0)(32.2)(4.33) / 0.33²

k = 2510 lbf / ft

giving a force of

F = 2510(0.33)

F = 836 lbf

or 26 times the weight of the block at rest.

One would probably have to run actual experiments to determine the actual force of impact because of the non uniform structure of a printer. This could give you an idea of the range of values to expect.

I hope this helps.

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Sorry you did not give enough information, Where are you dropping the 32 lb cinder block from, is it under water, in open air, or is it in a large vacuum tube within no air in it. all these things have a different effect on the dropping speed.

also what the 32 lb cinder block is dropped on has no effect on the equation.