Question

Urgent!!! Consider the mass-on-a-spring system as shown in the figure below. The spring has a spring constant of 1.45e+3 N/m, and the block has a mass of 0.807 kg. There is a constant force of kinetic friction between the mass and the floor of 4.68 N. Starting with the spring compressed by 0.101 m from its equilibrium position, how far will the block travel once it leaves the spring? (Assume that block leaves the spring at at the spring's equilibrium position, marked x=0 in the figure. Give your answer as the distance from the equilibrium position to the final position of the block.)

Answer 1

The block will travel a distance of 0.037 m.

Step-by-step explanation:

To find the distance the block will travel once it leaves the spring, we can use the principle of conservation of energy. Before the block leaves the spring, the potential energy stored in the spring is equal to the work done by friction. When the block leaves the spring, all of the potential energy is converted to kinetic energy. Using the equation for potential energy in a spring, we can find the distance the block will travel.

Applying the equation ΔPE = W_friction, where ΔPE is the change in potential energy and W_friction is the work done by friction, we have (1/2)kx^2 = fsd, where k is the spring constant, x is the compression of the spring, fs is the force of static friction, and d is the distance the block travels after leaving the spring. Solving for d, we get:

d = (1/2k)x^2/fs

Plugging in the given values, we have d = (1/2 * 1450 N/m * 0.101 m^2) / 4.68 N = 0.037 m. Therefore, the block will travel a distance of 0.037 m once it leaves the spring.