Hooke’s Law (± Hooke’s Law or Hookes Law)

Part A = k = 4.0*10^{4} N/m

Part B = x = 0.45m

Part C = No, typical small pickup truck springs are not large enough to compress 0.45 m.

Part D = The new springs should have a spring constant that is substantially larger than the spring constant of the old springs.

Solutions Below:

A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck.

Part A

^{-2}m. What is the effective spring constant of the spring system in the taptap?

We know that F = kx. Therefore:

65kg * g = -k(1.6*10^{-2})

-637N = -0.016k (the force is negative because gravity is downward)

k = 39,812,5

And to 2 significant figures:

k = 4.0*10^{4} N/m

Part B

First, solve for the total mass loaded onto the taptap:

m = (27 * 65) + (3 * 15) + (5 * 3) + 25

m = 1,840kg

And the gravitational force (the weight of everything loaded in the taptap) is:

F = ma

F = 1,840*9.8

F = -18,032 N (negative because the force is downward)

Using the k that we found in Part A, solve for x:

F = -kx

-18,032 = -4.0*10^{4}x

x = 0.45m

Part C

The answer is no. At least according to Mastering Physics:

No, typical small pickup truck springs are not large enough to compress 0.45 m.

Part D

New, stronger springs will compress less. So the spring constant k will have to be greater:

The new springs should have a spring constant that is substantially larger than the spring constant of the old springs.

For Part A, the power of 10 is supposed to be: 10^4 N/m

Yes indeed- the calculator figured it correctly but the written answer was wrong. Thanks for the correction!