Question

A 25.0 kg block slides down a ramp that is elevated at 36.0° a distance of 5.00 m. The coefficient of kinetic friction is 0.220. (a) What is the potential energy of the block before it begins to slide? (b) What is the work done by friction as the block slides down the ramp (said energy being converted into heat)? (c) What is the speed of the block when it reaches the bottom?

Please explain why we need to use sin 36 to calculate normal force. I know that we should only use cos 36 to calculate normal force.

Answer 1

a)

for the initial potential energy

initial potential energy = m * g * L * sin(theta)

L*sin(theta) is the height of the block from ground

initial potential energy = 25 * 9.8 * 5 * sin(36)

initial potential energy = 720 J

b)

work done by friction = -u * N * L

work done by friction = -u * m * g * L * cos(theta)

work done by friction = -0.220 * 25 * 9.8 * 5 * cos(36 degree)

work done by friction = -218 J

c)

let the speed at the bottom is u

Using work energy theorem

0.50 * 25 * u^2 = 720 - 218

solving for u

u = 6.34 m/s

the speed of block at the ground is 6.34 m/s