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Hotse xG thankex on wwwsxson www.xG A Due xen we xx on www.x XC Chege XC Tinal xS McNexte Mcne CO www.saplinglearningcom, ibiscms/mod/ibis/view.php?id=437391 A Sapling Learning Jump to. De'Angelo Vaughn-Allen macmillan learning Gradeboo # Attempts Score Assignment Information U) 10/20//2017 10:00 PM 79.6100 G O/19/2017 08:43 PM D Print A Calculator Perivcic Table 100 Queslin 5 of 20 Available From: 10/11/2017 10:0 Map | Due Date: 10/20/2017 10:0 A Sapling Learning Points Possible; 100 100 Grade Category: Graded A cart for hauling ore out of a gold mine has a mass of 443 kg, including its load. The cart runs along a straight stretch of track that is sloped 4.63" from the horizontal. A donkey, trudging along and to the side of the track, has the unenviable job of pul 1g the cart up the slope wi a 438-N force for distance of by means of a rupe that is parallel to the ground and makes an angle of 12.30 with the track. The coefficient of friction for the car's wheels on the track is 0.0177. Use g= 9.81 mis? Description: 100 Policies Find the work that the donkey perfoms on the cart during this process, 100 You can check your answers Number You can view solutions after the due 12.3 You can keep trying to answer each until you get it right or give up 100 track rope Find the work that the force of gravity performs on the cart during the process. There is no penalty for incorrect an For multiple-choice questions, ther depends on the number of choices 50 Number 100 O eTextbook 50 Calculate the work done on the cart during the process by friction. - - - "track incline Help With This Topic 100 horizontal ground There is a hint available! 100 O Web Help & Videos More that I dineon the divider bat again to hide the hit 100 Type here to search 9 a 4:13 PM 10202017 1Explanation / Answer
The work done by gravity is equal to the change in potential energy:
W done by gravity = -m*g*h
It's negative because gravity pulls in the opposite direction to the cart's displacement.
where h = (109 m) * sin(4.63°)
so:
W done by gravity = -(443 kg)*(9.81 m/s^2)*(109 m) *sin(4.63°)
W done by gravity = -3.82 * 10^4 J
Work done by they donkey equals the distance moved multiplied by the component of the force applied in the direction of movement:
W done by donkey = (109 m) * (438 N) *cos(12.3°)
W done by donkey = 4.6646 * 10^4 J
Work done by friction:
There are two ways to find the work done by friction. I'll do the hard way first...
First note that the work done "on the cart by friction" is negative. You know this for two reasons: First, friction tends to slow the cart down rather than speed it up, so it's taking energy from the cart (which is what doing negative work on something means). Secondly, you know that the force of friction is in the opposite direction of the cart's movement. So if you end up with a positive number, you know you made a mistake.
The force of friction is the coefficient of friction times the normal force between the cart and the track. Since the track is inclined at 4.63°, the normal force is mg*cos(4.63°). So the magnitude of the force of friction is:
F = -0.0177 * mg*cos(4.63°)
The negative sign is because force is in the opposite direction of displacement.
When the force is constant and the force is parallel to movement, work equals force times displacement.
W done by friction = F * (109 m)
W done by friction = -0.0177 * mg*cos(4.63°) * (109 m)
W done by friction = -0.0177 * (443 kg) * (9.81 m/s^2) *cos(4.63°) * (109 m)
W done by friction = -8.357 * 10^3 J
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