Question

The springs of a pickup truck act like a single spring with a force constant of (1.30 times 10⁵ N/m). By how much will the truck be depressed by its maximum load of (1000 kg)?

a) (7.0 cm)
b) (10.0 cm)
c) (12.0 cm)
d) (15.0 cm)

Answer 1

The truck will be depressed by approximately 7.54 cm.

Step-by-step explanation:

To find by how much the truck will be depressed, we can use Hooke's law which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. The formula is:

F = kx

Where F is the force exerted, k is the force constant, and x is the displacement.

In this case, the force constant k is given as 1.30 × 10^5 N/m and the maximum load on the truck is 1000 kg. We can convert this to Newtons by multiplying by the acceleration due to gravity (9.8 m/s^2):

F = (1000 kg)(9.8 m/s^2) = 9800 N

Now we can use Hooke's law to find the displacement:

9800 N = (1.30 × 10^5 N/m) * x

Solving for x:

x = 9800 N / (1.30 × 10^5 N/m) = 0.0754 m

Since the question asks for the depression in centimeters, we can convert the displacement to centimeters by multiplying by 100:

x = 0.0754 m * 100 = 7.54 cm

So the truck will be depressed by approximately 7.54 cm.