Hooke’s Law (± Hooke’s Law or Hookes Law)
Part A = k = 4.0*105 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.
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.
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*104 N/m
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*104x
x = 0.45m
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.
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.