Question

3. (20 Points) A cart of mass m is released from rest at the top of cireular quarter-anc ramp of radis R. The quarter-arc ramp of radius R. The friction on the ramp is negligible. The bottom of the ramp leads into a rough surface with a coeficient of kinetic friction no friction (a) (5 Points) What is the velocity of the cart at r -0 (b) (5 Points) What is the power of the frictional force on the cart at 0? (c) (5 Points) How far will the cart travel from the bottom of the ramp before stopping? (d) (5 Points) What is the work done by the frictional force in stopping the cart? (the start of the rough surface)?

Answer 1

a] Use conservation of energy

initially, all the energy is in the form gravitational potential energy

at the bottom, the energy is in the form of kinetic energy

so,

mgR = (1/2)mv²

=>

a] Use conservation of energy initially, all the energy is in the form gravitational potential energy at the bottom, the energy is in the form of kinetic energy so, mgR = (1/2)mv^2 v = squareroot 2 g R this is the velocity of the cart at x = 0. b] Average Power = force x velocity so, power of the frictional force = P = Frv P = mu m g squareroot 2 gR Acceleration offered by friction = a = -Fr/m a = -mu mg/m = -mu g initial velocity = u = [2gR]^1/2m/s final velocity = v = 0 m/s use, v2�= u2�+ 2aS 0 = 2gR - 2 mu gS S = R/mu this is how far the cart will travel horizontally. d] Work done by friction force in stopping the cart = change in kinetic energy of the cart W = -1/2 m[v^2 - u^2] W = 1/2 m[2 g R] = mgR

this is the velocity of the cart at x = 0.

b] Average Power = force x velocity

so, power of the frictional force = P = F_rv

=> $P = \mu mg \sqrt{2gR}$

c]

Acceleration offered by friction = a = - F_r/m

$a = - \frac{\mu mg}{m} = - \mu g$

initial velocity = u = [2gR]^1/2 m/s

final velocity = v = 0 m/s

use, v² = u² + 2aS

=> $0=2gR - 2\mu gS$

=> $S = \frac{R}{\mu}$

this is how far the cart will travel horizontally.

d] Work done by friction force in stopping the cart = change in kinetic enegy of the cart

$W = -\frac{1}{2}m[v^2-u^2]$

=> $W = \frac{1}{2}m[2gR] = mgR$ .