# Shouldn’t Work equal change in Kinetic Energy?

**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 -8t^{2}+4t^{3})i + (920 + 2t + 6t^{2} 5t^{3})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:

v_{x}(t) = derivative of (800-3t -8t^{2}+4t^{3}) = 12t^{2}-16t-3 pixels/s

v_{y}(t) = derivative of (920 + 2t + 6t^{2} 5t^{3}) = -15t^{2} +12t + 2 pixels/s

– – – Initial and Final Velocity:

v_{x}(0) = -3.0 pixels/s = – 0.6 m/s

v_{y}(0) = 2.0 pixels/s = 0.4 m/s

∴ v_{0} = √(0.4^{2}+0.6^{2}) = 0.72 m/s

v_{x}(1) = -7.0 pixels/s = -1.4 m/s

v_{y}(1) = -1.0 pixels/s = -0.2 m/s

∴ v_{1} = √(0.2^{2}+1.4^{2}) = √2

– – – Work Energy Theorem:

Work = ΔKE = (1/2)mv_{1}^{2} – (1/2)mv_{0}^{2}

Work = (1/2)(70)(2) – (1/2)(70)(0.72^{2})

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:

http://ift.tt/1f96zZV

## Leave a comment