Question

a 1210kg rollercoaster car is moving 7.33 m/s. as it approaches the station, breaks slow it down to 1.88 m/s over a distance of 5.29m. how much work did the break do?

Answer 1

\[x _{f}=x_{i}+\frac{ v _{f}-v _{i} }{ 2 } t \rightarrow t=2\frac{ x _{f}-x _{i} }{ v _{f}-v _{i} }=2\frac{ d }{ \Delta v }\]\[a=\frac{ v _{f}-v _{i} }{ t }=\frac{ \Delta v }{ t }\]\[F=ma\]\[W=Fd\]Therefore, \[W=m \frac{ \Delta v \Delta v}{ d }d=\frac{ 1 }{ 2 } m(\Delta v)^{2} =\frac{ 1 }{ 2 }1210kg(1.88m/s-7.33m/s)^{2}=17970.0125J\]You can also so it much simpler like this:\[W=\Delta E= E _{f}-E _{i}=\frac{ 1 }{ 2 }m \Delta v\]

Answer 2

We can use work energy theorem. Work done by ALL forces is always equal to change in the kinetic energy. We need not worry about calculating forces and acceleration. Or need not even know how much displacement the force(s) caused.

Find initial and final K.E. (You are given mass and initial and final velocities) and then simply find the change. The change in kinetic energy will be negative which means that work done by braking force is negative.

Answer 3

@ivancsc1996 bogus math alert!

\[\frac{1}{2}m v_2^2-\frac{1}{2}m v_1^2 \ne \frac{1}{2}m(v_2-v_1)^2\]

\((a-b)^2 = a^2-2ab+b^2, \text{ not } a^2-b^2\)

Answer 4

Oh yeah you are right, the which one is correcct? This the correct equation:\[W=\frac{ 1 }{ 2 }m (v _{f}^{2}-v _{i}^{2})\]

Answer 5

But where did I mess up in the first one I gave?

Answer 6

Yes, that's correct, but you equated
\[W=m \frac{ \Delta v \Delta v}{ d }d=\frac{ 1 }{ 2 } m(\Delta v)^{2} =\frac{ 1 }{ 2 }1210kg(1.88m/s-7.33m/s)^{2}=17970.0125J\]and there you squared the difference of the velocities rather than taking the difference of the squared velocities. There's a nonlinear relationship between velocity and energy, so you can't do that.

Answer 7

Yeah, I understand. What I don't get is why the algebra using the definition of W=Fd does not give out the same as W=Delta v. I just don't see what I did wrong in my algebra on the first equation

Answer 8

I am sure it is wrong, but were did I do it wrong

Answer 9

I know I missed a 1/2 at the final step of W.

Answer 10

The last line of your first post is also incorrect.
\[W=\Delta E= E _{f}-E _{i}=\frac{ 1 }{ 2 }m \Delta v\]\(E_f \) and \(E_i\) have units of \(\text{kg} \text{ m}^2/\text{s}^2\) but the expression on the RHS has units \(\text{kg}\text{ m}/\text{s}\)

Please don't suggest that it is just missing the ^2: \(m(\Delta v)^2 \ne m(v_f^2-v_i^2)\)

Answer 11

Yes you are right. I understand that. What I dont undesrtand is how the derivation before "You can also so it much simpler like this:" Is wrong. I understand the answer is wrong and I get that \[W= \Delta E=\frac{ 1 }{ 2 }mv _{f}^{2}-\frac{ 1 }{ 2 }mv _{i}^{2}=\frac{ 1 }{ 2 }m(v _{f}-v _{i})^{2}\neq \frac{ 1 }{ 2 }m \Delta v ^{2}\]It was a slip of the pen. I just don't understand how I derived the wrong formula from the equations on the first derivation. Please help me!!!!!!!!!!!!!!