(c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m.How much work does friction do on the sled?250XThe response you submitted has the wrong sign. JEXERCISEHINTS: GETTING STARTED I I'M STUCK!Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 55.0 N.(a) If he does 4.55 x 10² J of work on the sled while exerting the force horizontally, through whatdistance must he have pushed it?m(b) If he exerts the same force at an angle of 50.0° with respect to the horizontal and moves thesled through the same distance, how much work does he do on the sled?J PROBLEM An Eskimo returning from a successfulfishing trip pulls a sled loaded with salmon. The totalmass of the sled and salmon is 50.0 kg, and the Eskimoexerts a force of 1.20 x 10² N on the sled by pulling onthe rope. (a) How much work does he do on the sledif the rope is horizontal to the ground (0 = 0° in thefigure) and he pulls the sled 5.00 m? (b) How muchwork does he do on the sled if 0 = 30.0° and he pulls thesled the same distance? (Treat the sled as a pointparticle, so details such as the point of attachment of the rope make no difference.) (c) At a coordinateposition of 12.4 m, the Eskimo lets up on the applied force. A friction force .0 N between the ice andthe sled brings the sled to rest at a coordinate position of 18.2 m. How much work does friction do on thesled?SOLUTION(A) Find the work done when the force is horizontal.Use the proper equation, substitutingthe given values.7STRATEGY Substitute the given values of F and Ax into the basic equations for work.Use the proper equation, with Fxreplaced by fk²mgAn Eskimo pulling a sled with a rope at an angle to thehorizontal.(B) Find the work done when the force is exerted at a 30° angle.Use the proper equation, againsubstituting the given values.Substitute fk = -45.0 N and the initialand final coordinate positions into x;and xf.0W = F Ax = (1.20 x 10² N) (5.00 m) = 6.00 × 10² J(C) How much work does a friction force of 45.0 N do on the sled as it travels from a coordinate positionof 12.4 m to 18.2 m?W = (Fcos 0) d = (1.20 x 10² N) (cos 30.0°) (5.00 m)= 5.20 x 10² JWfric = FxAx= fk(xf = x;)Wfric= (-45.0 N) (18.2 m - 12.4 m)=-261 J

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Description: (c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m.How much work does friction do on the sled?250XThe re

A) Given - massM = 50 kg force F = 55 N…

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(c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m. How much work does friction do on the sled? 250 X The response you submitted has the wrong sign. J RCISE ose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 55.0 N. (a) If he does 4.55 x 10² J of work on the sled while exerting the force horizontally, through what distance must he have pushed it? HINTS: GETTING STARTED I'M STUCK! (b) If he exerts the same force at an angle of 50.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled? J

(c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m. How much work does friction do on the sled? 250 X The response you submitted has the wrong sign. J RCISE ose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 55.0 N. (a) If he does 4.55 x 10² J of work on the sled while exerting the force horizontally, through what distance must he have pushed it? HINTS: GETTING STARTED I'M STUCK! (b) If he exerts the same force at an angle of 50.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled? J

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Question

(c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of
50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m.
How much work does friction do on the sled?
250
X
The response you submitted has the wrong sign. J
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 55.0 N.
(a) If he does 4.55 x 10² J of work on the sled while exerting the force horizontally, through what
distance must he have pushed it?
m
(b) If he exerts the same force at an angle of 50.0° with respect to the horizontal and moves the
sled through the same distance, how much work does he do on the sled?
J

PROBLEM An Eskimo returning from a successful
fishing trip pulls a sled loaded with salmon. The total
mass of the sled and salmon is 50.0 kg, and the Eskimo
exerts a force of 1.20 x 10² N on the sled by pulling on
the rope. (a) How much work does he do on the sled
if the rope is horizontal to the ground (0 = 0° in the
figure) and he pulls the sled 5.00 m? (b) How much
work does he do on the sled if 0 = 30.0° and he pulls the
sled the same distance? (Treat the sled as a point
particle, so details such as the point of attachment of the rope make no difference.) (c) At a coordinate
position of 12.4 m, the Eskimo lets up on the applied force. A friction force .0 N between the ice and
the sled brings the sled to rest at a coordinate position of 18.2 m. How much work does friction do on the
sled?
SOLUTION
(A) Find the work done when the force is horizontal.
Use the proper equation, substituting
the given values.
7
STRATEGY Substitute the given values of F and Ax into the basic equations for work.
Use the proper equation, with Fx
replaced by fk²
mg
An Eskimo pulling a sled with a rope at an angle to the
horizontal.
(B) Find the work done when the force is exerted at a 30° angle.
Use the proper equation, again
substituting the given values.
Substitute fk = -45.0 N and the initial
and final coordinate positions into x;
and xf.
0
W = F Ax = (1.20 x 10² N) (5.00 m) = 6.00 × 10² J
(C) How much work does a friction force of 45.0 N do on the sled as it travels from a coordinate position
of 12.4 m to 18.2 m?
W = (Fcos 0) d = (1.20 x 10² N) (cos 30.0°) (5.00 m)
= 5.20 x 10² J
Wfric = FxAx
= fk(xf = x;)
Wfric= (-45.0 N) (18.2 m - 12.4 m)
=
-261 J

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