Final answer:
Calculating using Hooke's Law, the truck with a maximum load of 1000 kg and spring constant of 1.30 x 10^5 N/m will be depressed by 7.55 cm.
Step-by-step explanation:
The question is related to the physical concept of elastic deformation under the influence of a force. Particularly, it involves calculating the depression of the springs of a pickup truck when subjected to its maximum load. The force constant (k) of the spring is given as 1.30 × 105 N/m, and we have a maximum load (weight) which can be calculated from the mass (m = 1000 kg) using the equation weight = mass × gravitational acceleration (g). The gravitational acceleration (g) is approximately 9.81 m/s2.
To find the depression (x), we use Hooke's Law, which states that the force (F) exerted by a spring is equal to the negative product of its displacement (x) and its spring constant (k) (F = -kx). Solving for the displacement, we can express x as x = -F/k. Since we are looking for the magnitude of the displacement, the negative sign will be ignored in the final result.
So, given:
m = 1000 kg
k = 1.30 × 105 N/m
g = 9.81 m/s2
We can calculate the weight,
weight (F) = m × g = 1000 kg × 9.81 m/s2 = 9810 N.
Now, to calculate the depression (x):
x = F/k = 9810 N / 1.30 × 105 N/m = 0.0755 m or 7.55 cm.
Therefore, the truck will be depressed by 7.55 cm under its maximum load of 1000 kg.