51.6k views
3 votes
A 60.0-kg person drops from rest a distance of 1.20 m to a platform of negligible mass supported by an ideal stiff spring of negligible mass. the platform drops 6.00 cm before the person comes to rest. what is the spring constant of the spring?

2 Answers

6 votes

Final answer:

To find the spring constant, we use Hooke's law which states F = kx, where F is the force and x is the displacement. The weight of the person provides the force and the compression of the spring is x. Solving for the spring constant gives us k = 9800 N/m.

Step-by-step explanation:

The question involves finding the spring constant of a spring supporting a platform and a person. The energy stored in the spring due to the person's weight is equal to the potential energy lost by the person when they fell the 1.20 m distance onto the spring. To find the spring constant, we can use Hooke's law, which relates the force (F) to the spring constant (k) and the displacement (x): F = kx. In this scenario, the force applied on the spring is equal to the weight of the person (w = mg), which is countered by the spring force when the spring is compressed by 6.00 cm (0.06 m).

Considering the person's weight (mg) and the displacement of the spring (x), we can set up the equation mg = kx. We know the mass (m = 60.0 kg) and that g = 9.80 m/s² (acceleration due to gravity), so we can find the weight of the person is 60.0 kg × 9.80 m/s² = 588 N. Now we have F = 588 N and x = 0.06 m. We can solve for k using k = F/x, which gives us k = 588 N / 0.06 m = 9800 N/m as the spring constant.

User Patrick Kostjens
by
8.6k points
2 votes

Answer:4.12 *10^5

Step-by-step explanation:

2mgh/x^2

2(60)(9.8)(1.20)/(0.0600)^2= 4.12*10^5

User Gnat
by
7.3k points