Question

A 100 kg one-man bobsled starts from rest 50 m from the end of an icy, 15∘ slope. A giant 2.0-m -long spring is placed against a barricade at the bottom of the slope. (Figure 1) shows a coordinate system with the origin at the base of the spring. The bobsled brake fails to work, and the sled crashes into the spring. The spring compresses 75 cm before expanding and shooting the bobsled back up the slope. What is the spring constant of the spring?

Answer 1

The spring constant (k) is determined by equating gravitational and elastic potential energies. Substituting values yields k ≈ 1040 N/m for the spring after compressing 75 cm during the collision.

To find the spring constant, we can use the conservation of mechanical energy. Initially, the bobsled has gravitational potential energy that will be converted into elastic potential energy when compressed against the spring. The equation for gravitational potential energy is U_gravity = mgh, where m is the mass (100 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the vertical height of the slope.

The gravitational potential energy is then converted into elastic potential energy in the spring at the bottom of the slope. The elastic potential energy in a spring is given by U_spring = (1/2)kx^2, where k is the spring constant and x is the compression of the spring (converted to meters, so 0.75 m).

Setting these two energies equal, we get mgh = (1/2)kx^2. Solving for k, we find k = (2mgh)/(x^2).

Substitute the given values:

k = (2 * 100 kg * 9.8 m/s^2 * 15 m) / (0.75 m)^2

Calculating this gives the spring constant k.

The spring constant (k) is found to be approximately 1040 N/m.