Kinetic Energy and Work problem

In summary: That's why I said there was a mistake in the last calculation. (I should have said "the last calculation for phase 2".)
  • #1
**Mariam**
46
1

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 :)
 
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  • #2
Why are you using F = ma? Think carefully. Is there a better way to tackle this problem?
 
  • #3
**Mariam** said:
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.
 
  • #4
What mistake?
 
  • #5
**Mariam** said:
What mistake?
Oh the answer is 20 m/s.. But still that doesn't fix the final answer which is supposed to be 240
 
  • #6
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.
 
  • #7
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.
 
  • #8
**Mariam** said:
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.

**Mariam** said:
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.
 

What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is directly proportional to the mass and square of the velocity of the object.

How is kinetic energy calculated?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

What is the principle of conservation of energy?

The principle of conservation of energy states that energy cannot be created or destroyed, it can only be transferred or transformed from one form to another. In the case of kinetic energy, it can be converted into potential energy or other forms of energy, but the total amount of energy remains constant.

What is work in relation to kinetic energy?

Work is the transfer of energy from one object to another by applying a force over a certain distance. In the context of kinetic energy, work is done when a force is applied to an object to accelerate it and increase its velocity, thus increasing its kinetic energy.

Can kinetic energy be negative?

Yes, kinetic energy can be negative if the velocity of the object is negative. This means that the object is moving in the opposite direction of the applied force, resulting in a decrease in its kinetic energy. However, the magnitude of the negative kinetic energy is still equal to the magnitude of the positive kinetic energy in the opposite direction.

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