Marcus and Kiwi are going for a ride in their new wagon! Then they will roll down a frictionless hill, dropping a total height of 2.0 meters. How far into the leaf pile will the kitties and wagon travel before coming to a stop?

total info... Marcus and Kiwi are going for a ride in their new wagon! Marcus, Kiwi, and the wagon have a combined mass of 25 kg. Starting from rest, an impulse of 350 N*s will be exerted on the wagon and cats as they roll across horizontal, frictionless ground. Then they will roll down a frictionless hill, dropping a total height of 2.0 meters. At the bottom of the hill, the wagon will roll into a giant pile of leaves, exerting a constant force of 75N. How far into the leaf pile will the kitties and wagon travel before coming to a stop?

1 Answer
Steve J.
May 24, 2018

#d = 39.2 m#

Explanation:

Impulse has 2 definitions: forcetime and change in momentum. Remember that the Newton must be equivalent to the combination of units kgm/s^2 (because of F=ma). Therefore #N*s " is equivalent to "kg*m/s#.

So we can say that the impulse of 350 Ns will produce a momentum change of 350 kgm/s. Since they started from rest, the impulse gave them a momentum of #350 kg*m/s#. So their speed when they reach the top of the hill can be determined as follows:

#"momentum" = m*v = 350 kg*m/s#

Plugging in the mass and solving for v

#25 cancel(kg)*v = 350 cancel(kg)*m/s#

#v = (350 m/s)/25 = 14 m/s#

The kinetic energy due to their #14 m/s# speed is

#KE = 1/2*m*v^2 = 1/2*25 kg*(14 m/s)^2 = 2450 J#

At the top of that hill, they also have gravitational potential energy of
#GPE = m*g*h = 25 kg*9.8 m/s^2 * 2 m = 490 J#

After going down the hill, all their energy,

#2450 J + 490 J = 2940 J#,

will be kinetic energy.

The 75N retardation of the leaf pile will continue until they come to a stop. The 75N retardation does work until all 2940 J of kinetic energy have been converted to 2940 J worth of confusion in the leaf pile.

#"work" = F*d#

#"work" = 2940 J = 75 N*d#

#d = (2940 cancel(N)*m)/75 cancel(N) = 39.2 m#

I hope this helps,
Steve

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