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### Course: AP®︎/College Physics 1>Unit 4

Lesson 1: Work and mechanical energy

# Work-Energy Principle Example

The Work-Energy Principle tells us the amount of work done on an object or system will equal the change in kinetic energy for that object or system. Follow along in this worked example and understand about how to calculate the net work on an object and solve for the change in velocity. Created by David SantoPietro.

## Want to join the conversation?

• Just had a half-stupid question that I thought I might as well ask...
at In the last step of the video, David gets
``v(final)^2=81``
and equates that to
``v(final)=9``

So shouldn't the answer also equal `-9 m/s`? If so then how should we conceptualize this? Because anyways all our net forces/ velocities are in the same direction `(left)` which is considered +ve.

Also does this have anything to do with Work being a scalar and that -ve work is only a relative quantity?

Pls enlighten this dumb person...
• In the ideal world of Math, you're right, technically taking the square root would give you both a positive answer, and a negative answer.

However, in the case of Physics, as you also said, our velocity should be positive because this velocity has the same direction as the distance we've moved in.

If you went for a hike up a trail, trying to reach a mountain, going back down would be considered negative, as you're losing your hard-earned distance up. It's opposite to the distance you've covered.

If you're sledding down a hill, stopping your sled and going back up would be considered negative because you're going opposite to the distance you've been going.

I like to consider up positive and down negative, but whether or not your velocity is negative or positive is relative to your distance. It depends on if the velocity is going with your distance (+), or against your distance(-).

A velocity is positive (+) if that velocity is going in the same direction as our distance (d). ("Working with the distance is a positive thing to do!")

A velocity is negative (-) when that velocity is going in the opposite direction of our distance (d). ("Working against the distance is a very negative thing to do. >:/ ")

Looking at the problem, since there is a net force going left, and the distance we're going in is left, we know we should be speeding up. So our final answer of 9 would make more sense to be positive, going with our distance, as opposed to negative, which would mean we've started going backwards somehow.

Hope this helps!
• What does net mean in net force?
(1 vote)
• Net basically means the sum of all the forces. The same way net weight is the sum of all weights, the net force is the sum of all forces. Hope this helps!
• For those of you who are wondering at why `V_f` is POSITIVE 9 and not -9:

Note that the problem asked for the final speed not velocity. Remember back in the first lesson, Sal explained Speed is a Scaler Quantity, not a Vector Quantity like velocity. A Scaler Quantity only has a magnitude, no direction, so it's always NON-NEGATIVE.
• A 1,550 kg roller coaster cart is pulled at a constant speed of 4.3 m/s to the top of a
53.5 m tall ramp that has a 45 degrees incline. The coefficient of friction between the ramp and the roller coaster cart is μ=0.052. When the roller coaster cart makes it to the top of the ramp, the cable suddenly breaks and the roller coaster cart accelerates back down to the bottom of the ramp. How much work is done by friction?
I can understand the explanations for this practice question, but I don’t know what went wrong with my thinking:
When the cart is at the top, it has potential energy (KE=0), which is the transformed to KE as it rolls down the ramp. Therefore, we can use the work/energy principle. Epotential=change in KE=Wg + Wf. KE=mgh Wg=Fg*(square root of 2) / 2 h(square root of 2)=mgh. Then Wf = 0.
Intuitively this result doesn’t make sense when I know there is friction applied to the length of the ramp, but what is wrong with my thinking, please?
(1 vote)
• Note that there is friction applied throughout the length of the ramp with coefficient μ=0.052. This problem does not test the work-energy principle, and relies on the formula for work itself:

W_f = -fs = -(μmg cos(θ))*(2h/sin(θ)).

This formula will be covered later on in the course :)
(1 vote)