Problem 2: Suppose the ski patrol lowers a rescue sled carrying an injured skiier, with a combined mass of 97.5 kg, down a 60.0° slope at constant speed, as shown in the Figure. The coefficient of kinetic friction between the sled and the snow is 0.100. w 60° Otheexpert: Part (a) How much work, in joules, is done by friction as the sled moves 30.5 m along the hill? W; =- 1458.6 W = -1459 / Correct! Part (b) How much work, in joules, is done by the rope on the sled over this distance? W, = 2.38 * 104 W,= 23800 X Feedback: is available. Part (c) What is the work, in joules, done by the gravitational force on the sled? W = 2.53 * 104 W. = 25300 / Correct! Part (d) What is the net work done on the sled, in joules? W. 'net

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Chapter1: Units, Trigonometry. And Vectors
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Problem 2: Suppose the ski patrol lowers a rescue sled carrying an injured skiier, with a
combined mass of 97.5 kg, down a 60.0° slope at constant speed, as shown in the Figure. The
coefficient of kinetic friction between the sled and the snow is 0.100.
60°
©theexpertta.com
Part (a) How much work, in joules, is done by friction as the sled moves 30.5 m along the hill?
W; = - 1458.6
W; = -1459 vCorrect!
Part (b) How much work, in joules, is done by the rope on the sled over this distance?
W, = 2.38 * 104
W, = 23800 X
Feedback: is available.
Part (c) What is the work, in joules, done by the gravitational force on the sled?
W, = 2.53 * 104
W. = 25300
/ Correct!
Part (d) What is the net work done on the sled, in joules?
Wnet =
sin()
cos()
tan()
7
8
9
HOME
cotan()
asin()
acos()
E
4
5
6
atan()
acotan()
sinh()
1
2
3
cosh()
ODegrees O Radians
tanh()
cotanh()
END
VOl BACKSPACE DEL CLEAR
Submit
Hint
Feedback
I give up!
Transcribed Image Text:Problem 2: Suppose the ski patrol lowers a rescue sled carrying an injured skiier, with a combined mass of 97.5 kg, down a 60.0° slope at constant speed, as shown in the Figure. The coefficient of kinetic friction between the sled and the snow is 0.100. 60° ©theexpertta.com Part (a) How much work, in joules, is done by friction as the sled moves 30.5 m along the hill? W; = - 1458.6 W; = -1459 vCorrect! Part (b) How much work, in joules, is done by the rope on the sled over this distance? W, = 2.38 * 104 W, = 23800 X Feedback: is available. Part (c) What is the work, in joules, done by the gravitational force on the sled? W, = 2.53 * 104 W. = 25300 / Correct! Part (d) What is the net work done on the sled, in joules? Wnet = sin() cos() tan() 7 8 9 HOME cotan() asin() acos() E 4 5 6 atan() acotan() sinh() 1 2 3 cosh() ODegrees O Radians tanh() cotanh() END VOl BACKSPACE DEL CLEAR Submit Hint Feedback I give up!
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