Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 9

Question

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a = 50.0° slope at constant speed, as shown in Figure 6.22. The coefficient of friction between the sled and the snow is 0.100. Figure 6.22 (a) How much work is done by friction as the sled moves 30.0 m along the hill? Correct: Your answer is correct. J (b) How much work is done by the rope on the sled in this distance? Incorrect: Your answer is incorrect. J (c) What is the work done by gravity on the sled? Incorrect: Your answer is incorrect. J (d) What is the total work done? Incorrect: Your answer is incorrect. J Additional Materials

Explanation / Answer

here,

total mass, m = 90 kg
angle on inclined, A = 50 degrees

Frictional coefficient, u = 0.1

Component's of varois force acting,

Gravity,
Fgx = mgSinA = 90*9.8 * Sin50 = 675.651 N

Fgy = Normal Force = mg*CosA = 90*9.8*Cos50 = 566.939 N

Frictional Forece, Ff = u*normal Force = 0.1 * 566.939 = 56.694 N

From Newton Law of motion Fnet = 0

Fgx - T - Ff = 0 ( tension force)

T = Fgx - Ff
T = 566.939 - 56.939
T = 510 N

Work done = Force * Displacement

Part A:
displacement, d = 30 m

Work done = Ff * d
wf = 56.939 * 30
wf = 1708.17 J or 1.708 kJ

Part B:
wr = T*d
wr = 510 * 30
wr = 15300 J or 15.3 kJ

Part C:
Wg = Fgx *d
wg = 675.651 * 30
wg = 20269.53 J or 20.26 kJ

Part D:
Work = W_gravity + W_friction + W_rope
W = wg + wr + wf
w = 20269.53 + 15300 + 1708.17
w = 37.27 kJ

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