Page 4 of 6 ET9 9.P.069. Page 5 of 6 m-56.0-k9 person running at an initial spee
ID: 1794548 • Letter: P
Question
Page 4 of 6 ET9 9.P.069. Page 5 of 6 m-56.0-k9 person running at an initial speed of v - 4.50 m/s jumps onto an M - 112-k0 s to rest relative to the cart. The coefficient of kinetic friction between the person and Ant initially at rest (figure below). The person slides on the cart's top surface and finally cart is 0.360. Friction between the cart and ground can be ignored. (Let the positive be to the right.) the (a) Fi direction with the sign of your answer.) nd the final velocity of the person and cart relative to the ground. (Indicate the m/s (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (Indicate the direction with the sign of your answer) (c) How long does the friction force act on the person? (d) Find the change in momentum of the person. (Indicate the direction with the sign of your answer.) N S Find the change in momentum of the cart. (Indicate the direction with the sign of your answer.) N S (e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (Indicate the direction with the sign of your answer.) tn (f) Determine the displacement of the cart relative to the ground while the person is sliding. (Indicate the direction with the sign of your answer.)Explanation / Answer
given m = 56 kg
vi = 4.5 m/s
M = 112 kg
k = 0.36
a. final velocity of the person and the cart = vf
from conservation of moemntum
mvi = (m + M)vf
56*4.5 = (56 + 112)*vf
vf = 1.5 m/s
b. friction force acting on the person = k*mg ( while he is sliding)
f = k*mg = 0.36*9.81*56 = 197.7696 N
c. time for which friction force acts on the person = t
f = M*a
1.5 = a*t
t = 0.8494733 s
d. chang eof momentum of person = -56(4.5 - 1.5) = -168 kg m/s
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