3. An Eskimo is pulling a sled at constant speed across the frozen tundra. Becau
ID: 1786323 • Letter: 3
Question
3. An Eskimo is pulling a sled at constant speed across the frozen tundra. Because the sled is gaining no energy, the net-work done on the sled must be zero. The coefficient of kinetic friction between the sled and the snow is .-03. The sled has a mass of 145kg and the Eskimo pulls with a constant force at an angle of 30t as drawn. Review: The force diagram has been drawn for you in the picture Use the force diagram and Newton's second law to find the normal force of the ground on the sled. (Careful, the normal force here is NOT equal to the weight of the sled, mg) Use the normal force to find the friction force of the ground on the sled Find the force of the Eskimo's pull. a. b. c. d. d f orte ok Eskimo's pa is-Fe eFk0 135 .751 N Work/Energy: The Eskimo now let's go of the rope and friction brings the sled to a stop. Use work/energy to calculate how far the sled slides before stopping. Take the speed of the sled when the Eskimo lets go to be 3m/s.Explanation / Answer
along vertical
Fnet,y = 0
n + F*sintheta - mg = 0
normal force n = mg - F*sintheta
(c)
frictional force fk = uk*n = uk*(mg - F*sintheta)
(d)
speed is constant acceleration ax = 0
along horizontal
Fnet,x = m*ax
Fnet = 0
F*costheta - fk = 0
F*costheta = uk*(mg - F*sintheta)
F*cos30 = 0.3*(145*9.8 - F*sin30)
F*0.866 = 426.3 - 0.15 F
F*(0.15 + 0.866) = 426.3
F = 419.6 N
===============
work done by friction = -fk*x
from work energy relation work = change in KE
W = kEf - KEi
W = (1/2)*m*(vf^2 - vi^2)
vi = 3m/s
vf = 0
-0.3*((145*9.8) - (419.6*sin30))*x = (1/2)*145*(0-3^2)
stopping distance x = 1.79 m <<<<------ANSWER
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