Physics (plz help)

Question

A 2.0 kg package slides along a floor with speed of 4.0 m/s. It then runs into and compresses a

spring until the package momentarily stops. Its path to the initially relaxed spring is frictionless,

but as it compresses the spring, a kinetic frictional force from the floor, of magnitude 15 N, acts

on the package. If k = 10,000 N/m, by what distance d is the spring compressed when the

package stops?

Answer 1

Set the initial kinetic energy of the package equal to the sum of the elastic potential energy of the compressed spring and the absolute value of the work done by friction. It is sort of like using energy conservation, but with a friction term correction to account for lost energy.

m = mass of package (given)

v = initial speed of package (given)

k = spring constant (given)

f = magnitude of kinetic frictional force (given)

d = compression distance of spring = ? (same as stopping distance)

mv²/2 = kd²/2 + fd

kd²/2 + fd - mv²/2 = 0

This gives us a quadratic equation in d. Plugging in the values of k, f, m, and v gives

5000d² + 15d - 16 = 0

d ≅ 0.0551 meters or -0.0581 meters

Assuming the compression distance is positive, d ≅ 0.0551 meters