Problem 4: Energy and work done by friction A 62.0 kg skier...

Problem 4: Energy and work done by friction A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.20 m long between points A and B. The coefficient of friction between this patch and her skis is 0.30. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. A. What is the work done by friction on the rough patch? B. How fast is the skier moving (in m/s) when she gets to point B? C. How fast is the skier moving (in m/s) when she gets to point C? D. How much mechanical energy was lost by the skier in crossing the rough patch?

Transcript

00:01 Problem 7 .54, we have a 60 kilogram skier at the top of a slope that is 65 meters high, and we're told that friction acts on it as they descend the slope, and now we're asked how fast are they moving at the bottom of slope.

00:16 Okay, so here, assuming that they're not moving at the top, they only have potential energy, so they just have u1, and let's call that g, and then they have work acting on them as they move down the slope, so then they have work acting on them as they move down the slope, so then they have have work due to the friction acting on them.

00:35 And that's going to equal to the connect energy at the bottom.

00:38 So i'll write k2.

00:40 These are the positions, k1, k and then two.

00:45 So now we just need to plug in what we know, and we'll find what the speed is at the bottom.

00:51 So we have u1 is 60 times g times 65 meters.

01:05 And then we have worked through the friction is equal to negative one zero five zero zero joules and then k2 is just going to be one half m a square okay so plugging out all in we'll have 60 times g times 65 minus 10 500 that's kilogels.

01:43 And then that's equal to one half mass times velocity squared.

01:49 Okay, so then we know what the mass is.

01:51 So we can plug that in.

01:53 I didn't come in there because i missed it.

01:56 But then we should just get velocity equal to 30 .4 meters per second.

02:05 Okay.

02:06 So now that's part a.

02:09 Part b is now moving horizontally that's gear across.

02:12 This patch of soft snow where mu k is equal to 0 .2 we're told that it is 82 meters from this point to this point the average force of air resistance on skier is 160 newtons so we have a force is equal to 160 newtons it's acting against this gear as they're moving so it's pointed to the left into my diagram how fast are they going after crossing the patch.

02:51 So in this case, we're going to have to write down what each of the pieces are.

03:00 So we have initial connect energy.

03:01 So that's going to be again the k2.

03:05 Then we have the plus the work done by friction of the surface.

03:11 So i'll write down s for surface friction.

03:14 We have work done by the air friction.

03:19 So i wrote af.

03:22 And then that is equal to k, that's the kinetic energy at position 3.

03:33 Okay, so again, we need know what k2 is.

03:37 We already found what that is, or i'll just need to plug it in again...

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