Question

A box of mass M, starts from rest and slides down a frictionless hill of height h, and across a horizontal

floor of length L. This section of the floor of length L has coefficient of kinetic friction μk. The box then
hits the free end of a horizontal spring of force constant K and compresses the spring. The other end of the
spring is anchored firmly to a wall. The ground under the spring is frictionless. Take acceleration due to
gravity as g. All questions below require you to show work to receive full credit

(a) What is the speed v1 of the box when it arrives at the bottom of the frictionless hill in terms
of g and h?

Answer 1

The speed v1 of the box when it arrives at the bottom of the frictionless hill is given by v1 = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the hill.

Step-by-step explanation:

To find the speed v1 of the box when it arrives at the bottom of the frictionless hill, we can use the principle of conservation of energy. The potential energy gained by the box as it slides down the hill is equal to the kinetic energy it has at the bottom. The potential energy gained is given by mgh, where m is the mass of the box, g is the acceleration due to gravity, and h is the height of the hill. The kinetic energy at the bottom is given by (1/2)mv^2, where v is the speed of the box.

Equating the potential energy gained to the kinetic energy, we have mgh = (1/2)mv^2. Canceling out the mass m from both sides of the equation, we get gh = (1/2)v^2. Solving for v, we have v = sqrt(2gh).

Therefore, the speed v1 of the box when it arrives at the bottom of the hill is v1 = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the hill.