Question

A skateboarder with mass ms = 46 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.6 m/s. a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.

          A skateboarder with mass ms = 46 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.6 m/s.
a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.
b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.
c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.

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Added by Jose Antonio R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Adi S

07/28/2022

Step 1

The initial kinetic energy of the skateboarder is zero (since he starts from rest), and the final kinetic energy is 1/2*ms*vf^2. The work done by gravity is mgh, where g is the acceleration due to gravity. Show more…

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A skateboarder with mass ms = 46 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.6 m/s. a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.

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Transcript

00:01 Let us begin with subpart a.

00:05 The mass of the skateboarder is given as ms equals 46 kilograms.

00:12 The height of the ramp above the ground, that is hy, is given as 3 .3 meters.

00:20 The final velocity of the skateboarder is given as 6 .6 meter per second.

00:29 And the initial velocity, since it is starting from rest, we have it as 0 meter per second.

00:35 Now, using work energy theorem, we can write net work done.

00:42 Net work done equals change in kinetic energy.

00:47 We know that net work done is the work done due to gravity plus the work done due to friction.

00:54 Work done by friction equals change in kinetic energy is half m vf square, final velocity square, minus half m vf.

01:06 Sorry, v -i -square, which is the initial velocity square.

01:12 Now, from this expression, we can rearrange and obtain an expression for wf, which is the work done by friction.

01:20 This will be equals to half into m taken out as common.

01:25 Vf -square minus v -i -square into, sorry, minus w -g, which is the work done due to gravity.

01:35 Now we know that v -i is 0 and work done due to gravity is equals to m -g -h -y.

01:45 So here we can substitute for skateboarder.

01:48 The work done by the friction, w -f is equals to half into ms into vf -square.

02:00 V -f -square, v -i -square is 0, minus ms into g -y -h -y.

02:07 So this is the expression for work done by the friction.

02:14 This is the answer of subpart a.

02:19 Let us move on to subpart b.

02:22 Let us first draw a free body diagram to understand the problem.

02:28 So here we have a ramp and the skateboarder.

02:37 The skateboarder jumps on the skateboard and starts descending downwards...

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