Question

A 60.0 kg person running at an initial speed of 4.50 m/s jumps onto a 120 kg cart that is initially at rest. The person slides on the cart's top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.320. Friction between the cart and the ground can be neglected.

a.Find the final speed of the person and cart relative to the ground.

b.For how long does the frictional force act on the person?

c.Determine the magnitude of the displacement of the person relative to the ground while he is slipping on the cart.

d. Determine the magnitude of the displacement of the cart relative to the ground while the person is slipping on the cart.

Answer 1

a)

Conservation of Momentum. Momentum (P = m*v) must be constant:

P = 60.0 * 4.5 = 270 kg.m/s (or N.s)

v = 270/(120+60) = 1.5 m/s

b)                                                             

Fr = umg = 0.320 * 60.0 * 9.8 = 188.16 N

where g = gravitational constant. F = m*g,

and Fr = F * frictional constant.

F = ma

188.16 = 60.0 * a

a = -3.136 m/s^2

t = (1.5 m/s – 4.5 m/s ) / -3.136 m/s^2 = 0.957 s

c)

t = 0.957 s

a = -3.136 m/s^2

v0 = 4.5 m/s

d = v0*t + 1/2 a * t^2 = 4.3065 – 1.436

d = 2.87 m

d)

F = 188.16 N

m = 120 kg

a = F/m = 1.568 a

d = 0 + 1/2 at^2 = 0.718 m

g)

E = 1/2 mv^2

E0 = 0.5 * 60 * 3.4^2 = 346.8 J

Ef = 0.5 * (60 + 120) * 1.133^2 = 115.5 J

Change in Kinetic energy = 115.5 - 346.8 J

, = -231.3 J