# Homework Help: Mousetrap Calculations

1. Jan 31, 2016

### Tnguyen33

For my physics class, I did a mousetrap car and we were supposed to calculate some answer, and I would just like to know if I did my work right. Sorry, I know this is a lot of work.

The problem statement, all variables and given/known data

1. Calculate your mousetrap car’s average speed.
2. Calculate the fastest velocity of your racer over the 8 meter distance.
3. Calculate the average acceleration of the car as the string unwinds.
4. Calculate the net force acting on the car as it accelerates.
5. Calculate the frictional force (Fk) slowing the car down.
6. Calculate the coefficient of kinetic friction µk for wheels on the floor.
7. Using the net force from step #4, calculate the work done by the spring as the string unwinds.
8. Calculate the power generated by the car as the string unwinds.
9. Calculate the kinetic energy of the car at its highest velocity.
10. Calculate the work done by the frictional force to bring the car to a stop after it reaches its highest velocity from step #2.

Data:
total displacement=7.24m, total time=9s,
time the string unwind(acceleration)=7s, distance accelerated=4.72m,
time required to stop=2s, distance required to stop=2.52m
mass=0.170kg

The attempt at a solution

1. I just did the speed formula of distance/time giving me 0.804m/s
2. I solved for acceleration Δx=Vit+1/2at² (4.72=1/2a7²) so a=.193m/s².
Then I used the average acceleration formula for a=(Vf-Vi)/Δt to get 0.193=(Vf-0)/7 Vf=1.35m/s
3. Gotten from #2 a=.193m/s²
4. ∑F=MA →∑F=(0.170)(.193)→∑F=.03281N
5. I used the deceleration for this so a=(Vf-Vi)/Δt, I plugged in a=(0-1.35)/2 getting the deceleration of -0.675m/s²
Fk=MA→Fk=(0.170)(0.675)→Fk=0.115N
6. µk=Fk/Fn, Fn in this case would equal mass×gravity so µk=0.115/(0.170)(9.8) getting µk=0.069
7.∑F=Fspring-Fk→ .03281=Fspring-0.115 →Fspring=0.1478N.
Wspring=Fs×Δx→ Ws=0.1478×4.72 →Ws=0.698J
8. Pavg=∑W/t Since this work requires Fk, I included it getting Pavg=0.155/7→ Pavg=0.0221W
9. KE=1/2mv²→ KE=1/2×0.170×1.35²→ KE=0.155J
10. I believe that in this case Fk would be the same as Kinetic Energy as Fk need to stop that KE but Wk=Fk×Δx → Wk=0.155×2.52→ Wk=.391J

2. Jan 31, 2016

### Simon Bridge

The average speed is the total distance travelled divided by the total time spent on the journey - this is 7.24m/9s = 0.824m/s
When you show working for this sort of thing, make sure you say which distances and times you use.

... the question seems to suggest that you don't need the average acceleration to find the top speed.
Assuming constant acceleration in the accelerating stage, you can use $d=\frac{1}{2}vT$, check that number for the decelerating stage - since you know the distance travelled in each stage.
This formula is easily obtained from a v-t diagram, so you should sketch one.
From there use the average acceleration definition to get #3.

4 is OK. In 5, the "deceleration" is the negative of the acceleration ... you gave the deceleration as a negative number so the car must have been speeding up?
I don't think that is what you mean. It is porfectly sensible to say the car is accelerating when it is slowing down: it's the difference between the technical term and the common one.
For most of those, just check you used the correct numbers ...
10. Careful what you say: force and energy are different things so the force cannot be equal to the kinetic energy, which is what you said. The work done by friction, bringing the car to rest, is equal to the kinetic energy though ... well done.

I think you mostly just need to be a bit more careful.

3. Jan 31, 2016

### Tnguyen33

Thank you, that was exactly what I needed.