Shouldn't Work equal change in Kinetic Energy?

1. Mar 7, 2014

Jay520

1. The problem statement, all variables and given/known data

In a video game, a 70kg villain making an elusive maneuver flies across the screen according to the trajectory r = (800-3t -8t2+4t3)i + (920 + 2t + 6t2 –5t3)j in pixel units. If each screen pixel represents 0.2m in the world of the villain and t is time in seconds, how much work is done on the villain between t=0s and t=1s?

2. Relevant equations

Work = ΔKE

Velocity V = derivative of displacement with respect to time

3. The attempt at a solution

This should be solvable by setting Work equal to the change in Kinetic Energy. We can find Kinetic Energy because we know the mass of the villain, and we can calculate his change in velocity by differentiating his displacement functions. This is the method I used, but there must be an error somewhere because I keep getting the wrong answer.

- - - Velocity components as functions of Time:

vx(t) = derivative of (800-3t -8t2+4t3) = 12t2-16t-3 pixels/s

vy(t) = derivative of (920 + 2t + 6t2 –5t3) = -15t2 +12t + 2 pixels/s

- - - Initial and Final Velocity:

vx(0) = -3.0 pixels/s = - 0.6 m/s
vy(0) = 2.0 pixels/s = 0.4 m/s

∴ v0 = √(0.42+0.62) = 0.72 m/s

vx(1) = -7.0 pixels/s = -1.4 m/s
vy(1) = -1.0 pixels/s = -0.2 m/s

∴ v1 = √(0.22+1.42) = √2

- - - Work Energy Theorem:

Work = ΔKE = (1/2)mv12 - (1/2)mv02

Work = (1/2)(70)(2) - (1/2)(70)(0.722)

Work = 51.8 J

- - - -

This seems fine to me, but I'm told that this is the incorrect answer. What gives? I've literally spent the last three hours just thinking about this problem. I'm beginning to think the question is wrong. Here's a photo of the question:

https://d1b10bmlvqabco.cloudfront.n...d5/hq5yd6sorm16ob/hsi4gvobc5tk/piazza_new.png

Last edited: Mar 7, 2014
2. Mar 7, 2014

Staff: Mentor

It looks like the acceleration depends on time, so I would think you need to integrate over the path...

3. Mar 7, 2014

Simon Bridge

You got 51.8J, which is one of the options - in fact it's the favorite!
The correct answer is given as 3.35J.

That's what I was thought at first.

looking at a 1D example:

$x(t)=t^3\\ v(t)=3t^2$

for 1kg mass:

$KE(t=0)=0\text{J}\\ KE(t=1)=4.5\text{J}$
suggests work is: $\Delta KE = +4.5\text{J}$

Lets check the path integral:

$a(t)=6t\\ F(t)=ma(t)=6t$

$$W=\int_{x_0}^{x_1} F(x)\;\text{d}x = \int_{t_0}^{t_1} F(t)\dot x \;\text{d}t = \int_0^1 (6t)(3t^2)\;\text{d}t=18\int_0^1 t^3\;\text{d}t=\frac{18}{4}=4.5\text{J} = \Delta KE$$
... but it may be worth checking.

How could a 51.8J change in KE happen with only 3.35J of work?
Not found a mistake in the algebra.

Last edited: Mar 8, 2014
4. Mar 8, 2014

Dick

Sure. In general, $\int F dx = \int ma dx =m \int \frac{dv}{dt} dx = m \int \frac{dx}{dt} dv = m \int v dv = m \frac{v^2}{2}$. Must be a mistake the the solution.

5. Mar 8, 2014

Simon Bridge

- it's either that or there's a context missing.
Looks like a past exam paper - it should have been picked up before now.
Considering the popularity of the choice D, it may be worth mentioning to the appropriate people.