PHYSICS: I need HELP with physics, someone please help a.s.a.p. Greatly appreciated thanks!?

Question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected with a winch. The shaft is inclined at 30.0 degrees above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 seconds and then continues at a constant speed. (a) What power must the winch motor provide when the car is moving at a constant speed? (b) What maximum power must the motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of lenght 1,250 m?

Answer 1

Power is the rate at which work is done. That is:

Power = work / time

Work is force times distance:

work = force × distance

Therefore:

Power = force × distance / time

Since distance/time = speed, you could also say:

Power = force × (distance/time)
Power = force × speed

> (a) What power must the winch motor provide when the car is moving at a constant speed?

Power = force × speed

They tell you that the speed is 2.20m/s, so you have that part.

The force equals how hard the winch is pulling. Since the car is no longer accelerating, the NET force on the car must be zero (zero acceleration ALWAYS means zero net force). That means the winch is pulling upslope exactly as hard as gravity is pulling downlsope; making the two opposing forces exactly cancel out.

Gravity always pulls downslope with a strength of (mg)sinθ (in this case, m=950kg and θ=30 degrees).

That means (since no acceleration), the winch's force exactly matches that in the opposite direction. Therefore:

Power = (force of winch) × speed
Power = (mg)sinθ × 2.20m/s
[During constant-speed phase]

> (b) What maximum power must the motor provide?

Power = force × speed

The maximum power happens when the force is maximum and the speed is maximum.

The maximum force (by the winch) happens while the car is _accelerating_. When the car is NOT accelerating, the winch's force is EQUAL to gravity; but in order to make the car _accelerate_, the winch must pull HARDER than gravity.

We can use the formula Fnet=ma, to determine how hard the winch is pulling during the acceleration.

The net force is:
Fnet = (F_winch − F_gravity)

Remember that F_gravity (downslope) is always: (mg)sinθ; so:

Fnet = (F_winch − (mg)sinθ)

Set this equal to "ma":

Fnet = ma
(F_winch − (mg)sinθ) = ma

Now figure out how much "a" is. Acceleration equals change in velocity divided by time. They tell you that the car took 12 seconds to go from zero to 2.20m/s; so:

a = Δv/t
a = (2.20m/s)/(12s)
a = 0.1833 m/s²

Combine with previous equation to get:

(F_winch − (mg)sinθ) = m(0.1833 m/s²)

Solve for F_winch:

F_winch = m(0.1833 m/s²) + (mg)sinθ

This is the maximum force provided by the winch.

Then:

(max power) = (max force) × (max speed)
(max power) = (m(0.1833 m/s²) + (mg)sinθ) × (max speed)
(max power) = (m(0.1833 m/s²) + (mg)sinθ) × (2.20m/s)

> (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of lenght 1,250 m?

One way to calculate this is to figure out the work done during both phases (the acceleration phase and the constant-speed phase) and then add those two work amounts together. That would require you to figure out the distance traveled during each phase too.

But that's too hard. The easier way, is to notice how high the car got lifted against gravity. That height is:

h = (total length of slope) × sinθ

Whenever you lift something by a height of "h", and providing there's no friction, the amount of work done on it is:

total work = (weight) × (h) = mgh

So:

total work = mg × (total length of slope) × sinθ

Answer 2

Wow, Thankss! I was wondering the same thing yesterday