A toy sled of mass 0.250 kg is shot from a catapult with an initial velocity of

Question

A toy sled of mass 0.250 kg is shot from a catapult with an initial velocity of 0.725 m/s. The coefficient of kinetic friction between the sled and the floor is 0.450. How far does the sled move before coming to rest? Use the concepts of work and kinetic energy to solve the problem.
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A boy riding a bicycle (total mass 72.0 kg) at a speed of 20.0 m/s collides head-on with another boy on a bicycle (total mass 85.0 kg) traveling in the opposite direction at the same speed. If the bicycles remain locked together immediately after the collision, how fast are they moving, and in which direction? How much kinetic energy has been lost in the collision?

Explanation / Answer

1. Energy in = Energy out KE = Work of friction .5mv^2 = umgd .5(.250)(.725^2) = .450(.250)(9.8)d d= .0595946m 2. P in = P out mv + mv = (m + m)v 72(20) + 85(-20) = (72 + 85)v v= -1.7105 m/s in the initial direction of the 85 kg boy KE lost = KE initial - KE after KE lost = (.5mv^2 + .5mv^2) - .5mv^2 KE lost = [.5(72)(20^2) + .5(85)-20^2)] - .5(72 + 85)(-1.7105^2) KE lost = 31170.316 joules

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