11. (18 pt) The figure shows a box of weight w = 2.0N...

11. (18 pt) The figure shows a box of weight w = 2.0N sliding down along the inclined plane from rest. The vector of displacement of the block is d=1.8 m. The plane makes an angle of 30° with the horizontal. The normal force acting on the box has a magnitude n = 1.7N, the coefficient of kinetic friction between the box and the plane is ?k = 0.30. A) What is the work Ww done on the box by the weight of the box? B) What is the work Wn done on the box by the normal force? C) What is the work Wfk done on the box by the force of kinetic friction? D) What is the kinetic energy of the box after it slides down 1.8 m? E) What is the change of gravitational potential energy of the earth-box system after it slides down 1.8 m? F) What is the change of thermal energy of the box-ramp system? G) Identify an isolated system in this problem and sketch the work-energy bar graphs of the system when the box is at the bottom and top of the ramp.

Transcript

00:01 A 2 -newton block slides down from rest from the top of the ramp and the displacement is 1 .8 meters.

00:08 Angle of inclination of the ramp is 30 degrees and there's a coefficient of kinetic friction between the ramp and the box at 0 .30.

00:18 We are given the normal force acting on the box and that is 1 .7 newton.

00:23 So if we show the free body diagram, this would be the normal force and this would be the weight exerted.

00:31 By the earth and there is kinetic friction that is opposing the direction of motion so w here we'll have let's show the components of w this would be w x if we take the ramp as the x -axis and there's this w y if we take the axis along the normal force as the y and the angle between the weight and the y axis is also theta there are several questions here for us to answer.

01:03 Let's start with the first one.

01:05 What is the work done by the weight? okay.

01:09 So you know in your lecture that work happens if and only if the applied force is along the axis where the change in position happens.

01:21 So for the weight, wy is perpendicular to the displacement, hence w sub y will do no work and it's only wx that will do work.

01:30 Since the direct.

01:31 Since the of displacement is downward and w sub x is also downward we are expecting that the work done by the weight is positive so we just put here the work done by weight is equal to the component along the direction of displacement that's w x and then of course times the magnitude of the displacement okay w x is just the weight which is two times uh sign angle 3 .000 theta because wx is opposite to angle theta.

02:07 This is wx.

02:09 And then times distance 1 .8 meters.

02:12 Therefore, let's move this higher.

02:14 Therefore, the work done by the weight is positive 1 .0 in two significant figures.

02:23 Jules.

02:24 Let's go to letter b.

02:26 Work done by the normal force.

02:28 So immediately we know that the work done by the normal force.

02:32 Force is zero.

02:34 Since the normal force is 90 degrees all throughout the motion, it's 90 degrees with the displacement vector.

02:42 Okay, that's it.

02:45 Cosine of angle 90 is zero.

02:48 Next, what about the work done by friction? okay, work done by friction.

02:52 Friction is a non -conservative force, and it removes energy from a system.

02:59 So it always makes a negative work because it's taking energy from the system.

03:03 This is just negative times the kinetic friction times the distance traveled.

03:11 So kinetic friction is just equivalent to mu k times the normal force times d.

03:17 In this case, the normal force magnitude is equivalent to w sub y.

03:23 So this is w cosine theta.

03:27 Let's further simplify this.

03:31 So we have negative mu k.

03:33 Normal force is weight, cosine theta, times the distance, negative 0 .30 times 2 newtons, times cosine 30 degrees times 1 .8.

03:49 Therefore, the work done by friction is negative 0 .94 jules.

03:58 Okay, that else.

04:03 Letter d, what is the kinetic energy of the block at the bottom of the ramp? so we label it as 0 .2 here.

04:10 We can use the work energy theorem, which says that the total work done on a system is equivalent to the change in its kinetic energy.

04:21 Okay, it has no speed initially, so this is zero.

04:25 And the total work here, we just need to add the work done by the weight plus the work done by friction...

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