Notes Ask Your Teacher ·I-10 points SerPSET9 9 P069 An m = 56.0-kg person runnin

Question

Notes Ask Your Teacher ·I-10 points SerPSET9 9 P069 An m = 56.0-kg person running at an initial speed of v = 4.50 mSjumps onto an M-112-kg cart initially at rest (figure below) on the cart's top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between and the cart is 0.360. Friction between the cart and ground can be ignored. (Let the positive direction be to the right.) The person slides (a) Find the final velocity of the person and cart relative to the ground. (lindicate the direction with the sign of your answer) 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. (Tndicate the direction with the sign of your answer.) (t) Determine the displacement of the cart relative to the ground while the person is sliding. (Indicate the direction with the sign of your answer) (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (0 Explain why the answers to (9) and (h) differ. (What kind of collision is this one, and what accounts for the loss of mechanical energy?)

Explanation / Answer

given, m = 56 kg

v = 4.5 m/s

M = 112 kg

k = 0.36

a. let final speed of the cart be u

then from conservation of meometnum

mv = (m + M)u

u = 1.5 m/s

b. friction force on the person = k*mg = 0.36*56*9.81 = 197.7696 N

c. acceleration of the person = f/m = 3.5316 m/s/s

hence

from v = 4.5 m/s to 0 m/s in time t

4.5 = 3.5316*t

t = 1.2742 s

d. change in momentum of the person = -56*(4.5 - 1.5) = -168 kg m/s

change in momentum of the cart = -change in momentum of th eperson = 169 kg m/s

e. displacement of person relative to ground while he is sliding on the cart = x

then

x = 4.5*1.2742 - 0.5*3.5316*1.2742*1.2742 = 2.866972476888 m

f. displacement of cart while the person is sliding = x'

x' = 0.5*(197.7696/112)*1.2742^2 = 1.433463761556 m

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