# Friction and Power Problem

1. Oct 16, 2006

### needhelp83

What is the minimum work needed to push a 950-kg car 310 m up a 9.0oincline?
a) ignore friction,
b) assume the effective coefficient of friction is 0.25

a)
Fnet=mg sin =(950 kg) (9.8 m/s2)(sin 9°)= 1456 N

W=Fd
W=(1456 N)(310 m)= 451,360 J

b)
Fnet=mg sin - µ mg cos
Fnet=(950 kg) (9.8 m/s2)(sin 9°)- (0.25)(950 kg) (9.8 m/s2)(cos 9°)=
Fnet=1456 N-2299 N=-843 N

W=Fd
W=(-843 N)(310 m)= -261330 J Should this be negative?

In a rope climb, a 70-kg athlete climbs a vertical distance of 5.0 m in 9.0 s. What minimum power output was used to accomplish this?

New to power problems? How should I set this up?

2. Oct 16, 2006

### needhelp83

Anybody?? :)

3. Oct 17, 2006

### andrevdh

a) fine
b) using the work-energy relation we have that the work done by the applied force and friction will gives us the change in the mechanical energy of the car:

$$W_a + W_f = \Delta E_m$$

the change in the mechanical energy can be taken to consist of only the change in potential energy of the car if we start out of rest and end in rest. The work done by friction will be negative so we can transfer the positive of this to the other side of the equation. So you can see that you need to add the raise in potential energy and the positive of the work done by friction.

For the rope climb problem you need to divide the amount of work done by the climber by the time taken to do that amount of work to get the power - it is the rate at which he/she is doing work in watt.