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adelina 88 [10]
3 years ago
9

Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loa

ded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force on the sled by pulling on the rope. Suppose the coefficient of kinetic friction between the loaded sled and snow is 0.200.
(a) The Eskimo pulls the sled 5.90 m, exerting a force of 1.10 102 N at an angle of θ = 0°. Find the work done on the sled by friction, and the net work. Wfric = Correct: Your answer is correct. . J Wnet = Correct: Your answer is correct. . J
(b) Repeat the calculation if the applied force is exerted at an angle of θ = 30.0° with the horizontal. Wfric = J Wnet = J
Physics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

(a)

W_friction = 98.1 J

W_net = 550.9 J

(b)

W_friction = 98.1 J

W_net = 463.95 J

Explanation:

(a)

First, we will calculate the work done by friction:

W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\

<u>W_friction = 98.1 J</u>

Now, the work done by Eskimo will be:

W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 0^o\\

W_Eskimo = 649 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 649 J - 98.1 J

<u>W_net = 550.9 J</u>

<u></u>

(b)

First, we will calculate the work done by friction:

W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\

<u>W_friction = 98.1 J</u>

Now, the work done by Eskimo will be:

W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 30^o\\

W_Eskimo = 562.05 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 562.05 J - 98.1 J

<u>W_net = 463.95 J</u>

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