Question

If a 2.6 kilograms can of cranberry sauce with the radius of one point five meters is released from rest at the top of 15 mile high hill. if the hill is frictionless what is the velocity of the can's center of mass at the bottom of the hill?

Answer 1

To solve this problem, we can use conservation of energy. At the top of the hill, the can of cranberry sauce has gravitational potential energy given by:

U = mgh

where m is the mass of the can, g is the acceleration due to gravity, and h is the height of the hill. We can plug in the given values to get:

U = (2.6 kg)(9.81 m/s^2)(15 miles x 1609.34 m/mile) = 601266.8 J

At the bottom of the hill, all of this potential energy will be converted into kinetic energy:

K = (1/2)mv^2

where v is the velocity of the can's center of mass at the bottom of the hill. We can solve for v by equating K and U:

(1/2)mv^2 = mgh

Simplifying and solving for v, we get:

v = sqrt(2gh)

Plugging in the given values, we get:

v = sqrt(2 x 9.81 m/s^2 x 15 miles x 1609.34 m/mile) = 423.6 m/s

Therefore, the velocity of the can's center of mass at the bottom of the hill is 423.6 m/s.