Doncouse_Jeremy_Week2

# Doncouse_Jeremy_Week2 - Running head WEEK 2 1 1 Week 2...

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Running head: WEEK 2 1 1 Week 2 Assignment Jeremy Doncouse Mountain State University

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Week 2 2 2 Week 2 Assignment Questions - Phys 101 Online Energy and Conservation Laws 1. A runner with a mass of 60kg accelerates from 0 to 9 m/s in 3 s. Find the net force on the runner using the alternate form of Newton’s second law. F = MA = M(ΔV/ΔT) where F is the force, M is mass, A is acceleration, ΔV is the change in velocity, and ΔT is the change in time. F = 60kg ( 9m/s / 3s) = 180 N 2. A 4000kg truck runs into the rear of a 1000kg car that was stationary. The truck and car are locked together after the collision and move with speed 10m/s. What was the speed of the truck before the collision? Compute how much kinetic energy was “lost” in the collision? a) First we must use momentum conservation to figure out the cars speed before collision. M1V1 + M2V2 = (M1 + M2) V3 where M1 is the mass of car, M2 is the mass of truck, V1 is the velocity of car prior to collision, V2 is the velocity of truck prior to collision, and V3 is the velocity of both locked vehicles. Since the car was stationary, V1 = 0 Therefore: M2V2 = (M1 + M2) V3 4000kg V2 = (4000kg + 1000kg) 10m/s V2 = 12.5m/s b) Now that we have the velocities prior to the collision, we can calculate the KE’s KEbefore = ½ M2V2 = ½ 4000kg (12.5m/s)2= 312500J KEafter = ½ (M1+M2) V3 = ½ (4000kg + 1000kg)(10m/s)2 = 250000J KEchange = KEafter – Kebefore = 250000J – 312500J = -62500J 3. Two persons on ice skates stand face to face and then push each other away.
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## This note was uploaded on 11/27/2011 for the course MATH 116 taught by Professor Unknown during the Spring '09 term at Mountain State.

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Doncouse_Jeremy_Week2 - Running head WEEK 2 1 1 Week 2...

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