- #1
**Mariam**
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Homework Statement
[/B]
A horizontal force of 50 N is applied to a 2.0 kg trolley, initially at rest, and it moves a distance of 4.0 m along a level, frictionless track. The force then changes to 20 N and acts for an additional distance of 2.0 m.
(a) What is the final kinetic energy of the trolley?
(b) What is its final velocity?
How do you solve this question?
2. The attempt at a solution:
First i divided the problem into two phases the first is where the force is 50 N and the second where it is 20. Then I found the acceleration for each which enabled me to find final velocity by kinematics equations.
phase 1:
F=ma... 50=2*a...a=25 m/s^2
vf=√(25*2*4)=10√2 m/s
K.E=1/2mv^2...1/2*2*(10√2)^2=50J
phase 2:
F=ma... 20=2a... a=10m/s^2
vf=√(2*10*2)=2√10 m/s
K.E=1/2mv^2... 1/2(2)(2√10)^2=40J
then i added the energy to find final K.E=90 J
however my book says its 240J
for final v it is 2√10=6.32
but book says 15m/s
What am I doing wrong and what's the correct way to solve this
Thanks :)