Kinetic Energy and Work problem

[**Mariam**]

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 :)

 

[PeroK]

Why are you using F = ma? Think carefully. Is there a better way to tackle this problem?

 

[Doc Al]

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

There's nothing wrong with this approach, except for a mistake in the last calculation.

But the easy way to find the energy (and then the speed) is to use the work-energy theorem. And, given the title of this thread, I suspect that that was what they want you to use.

 

[**Mariam**]

What mistake?

 

[**Mariam**]

What mistake?

Oh the answer is 20 m/s.. But still that doesn't fix the final answer which is supposed to be 240

 

[gneill]

Recheck your math for the first KE.

There's a much easier way to solve this if you realize that the work done yields the change in KE, and the work done by a force is just F*d when the force and displacement are aligned.

 

[**Mariam**]

Ok so I used the work energy theorem here is what i got for final velocity:

W=ΔK.E
Fd=1/2mvf^2-1/2mvi^2
(20)(2)=1/2(2)(vf)^2-1/2(2)(14.2)^2... (10√2=14.2 is vi and i got it from phase 1)
solving for vf... vf=15.5 m/s

and final K.E we just substitute vf in K.E equation... K.E=240.25 J

thanks it turned out to be easier than i thought actually.

 

[Doc Al]

Ok so I used the work energy theorem here is what i got for final velocity:

W=ΔK.E
Fd=1/2mvf^2-1/2mvi^2
(20)(2)=1/2(2)(vf)^2-1/2(2)(14.2)^2... (10√2=14.2 is vi and i got it from phase 1)
solving for vf... vf=15.5 m/s

Even easier: to find the final energy just add up the work done in each phase.

But still that doesn't fix the final answer which is supposed to be 240

That's because you messed up phase 2 of your first approach. Phase 2 has an initial non-zero velocity, which you did not take into account properly.