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Two teenagers pull a 19 kg child in a 3.7 kg sled up an incline of 10 degree wit

ID: 1563752 • Letter: T

Question

Two teenagers pull a 19 kg child in a 3.7 kg sled up an incline of 10 degree with ropes. Both teenagers pull with the same force at an angle of 30 degree relative to the forward direction (see the figure). The coefficient of static friction between the snow and the sled is 0.4 and the coefficient of kinetic friction is 0.25. (a) What is the minimum force each teenager should apply to get the sled moving in the forward direction? (b) If the teenagers continue to apply the same force after the sled is in motion, what is the acceleration of the sled? (c) If the slope is 20 m long, and the sled starts at rest, how long does it take them to get to the top of the slope? (d) What is the speed of the sled at the top of the slope? (e) What is the minimum force required to keep the sled in motion with a constant velocity?

Explanation / Answer

here,

mass of slide + child , m = 19 + 3.7 = 22.7 kg

theta = 30 degree

us = 0.4 and uk = 0.25

a)

the minimum force applied by each teenager be f

2 * f * cos(30) = us * m * g * cos(10)

2 * f * cos(30) = 0.4 * 22.7 * 9.81 * cos(10)

f = 50.65 N

b)

for moving sled, kinetic friction is used

the accelration , a = net force/effective mass

a = ( 2 * f * cos(30) - uk * m * g * cos(10)) /m

a = ( 2 * 50.65 * cos(30) - 0.25 * 22.7 * 9.81 * cos(10)) /22.7

a = 1.45 m/s^2

the accelration is 1.45 m/s^2

c)

l = 20 m

the time taken to reach the top be t

l = 0 + 0.5 * a * t^2

20 = 0 + 0.5 * 1.45 * t^2

t = 5.25 s

d)

the speed of the sled at the top , v = 0 + a * t = 7.62 m/s

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