How Is Work Related to Changes in Kinetic Energy in a Rocket Car's Deceleration?

[mousemouse123]

A rocket car is traveling at 648km/h[f] when the parachute is deployed and does 51.64MJ of work to slow the car down to a speed of 54.o km/h

Homework Equations

w = delta ek
w= m(1/2mvf^2 - 1/2mvi^2)
m= w/0.5(Vf^2-Vi^2)

The Attempt at a Solution

i derived that equation from w=delta kinetic energy and i got 2/3 for showing work... this was a test question 4 months ago

now i have an exam tomorrow.

the mass i got is 248kg. the teacher took it up and the correct mass was 3210kg. i honestly feel my answer is correct and i constantly repeat this question and get 248 kg!

he said my formula is correct just the calculation is wrong

this is what i did.

51.64x10^6J/ -208494(negative since Vi > Vf so you get negative mass but i guess you ignore the negative)

can anyone tell me what i am doing wrong?!

 

[gneill]

Recheck your value for v₀² - v_f².

 

[verty]

If the velocities were 648 m/s and 54 m/s, the answer would be 248 kg.

Let 1 km/h = c m/s

m = 2w / (v_1^2 - v_2^2)
= 2 * 51.64 * 10^6 / (648^2 c^2 - 54^2 c^2)
~= 248 / c^2

This formula only works for velocities in m/s, we see.

 

[mousemouse123]

If the velocities were 648 m/s and 54 m/s, the answer would be 248 kg.

Let 1 km/h = c m/s

m = 2w / (v_1^2 - v_2^2)
= 2 * 51.64 * 10^6 / (648^2 c^2 - 54^2 c^2)
~= 248 / c^2

This formula only works for velocities in m/s, we see.

i am no expert in physics but i think you made mistake. shouldn't it be 54^2-648^2 not the way you wrote it since it is slowing down. you subtracted v initial from v final... the formula is v final minus vinitial

 

[verty]

i am no expert in physics but i think you made mistake. shouldn't it be 54^2-648^2 not the way you wrote it since it is slowing down. you subtracted v initial from v final... the formula is v final minus vinitial

w = \Delta E_k = (1/2) m (v_f^2 - v_i^2)

I should have made the work negative because the change of energy was negative, energy was lost. So I made two changes to the formula, I neglected to make the work value negative, but I also wrote the velocities back to front. I could do that for this formula because it made no difference.

You can see that because:

\frac{2w}{v_f^2 - v_i^2} = \frac{2(-w)}{v_i^2 - v_f^2}.

So I actually used the second part of that, because I used -w (by using 51MJ and not -51MJ). Admittedly, I should have used -51 MJ and done it the correct way round, just to be safe.

I only did it because I could do it without making the answer invalid. But as you saw, what I could not do was use km/h values for the velocities.