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## Homework Statement

We're told to find the coefficient of friction for a mousetrap car that we've made, using values that we attained ourselves. We released the arm of the mousetrap onto a scale to find Fa → 1.0*10^2 Newtons.

This is how we found

**Fa**:

m = 1.020kg (←mousetrap held back and hammer released onto scale, value recorded from scale)

a = 9.8m/s/s (←factoring in gravity, which affects mousetrap's "release weight" on scale)

Fa = ?

Fa = ma

= 1.020kg * 9.8m/s/s (Fg)

= 9.996 Newtons

**(← rounded to correct amount of sig digs.)**

__= 10. Newtons__Using that, we solved for

**acceleration**:

Vi = 0.0m/s

t = 1.56s (time to complete 3m)

d = 3.0m

a = ?

d = Vi(t)+0.5a*(t^2)

a = (3m)/(0.5)(1.56^2)

a = 2.46 m/s/s

**(← rounded to correct amount of sig digs.)**

__= 2.5m/s/s__With that

**Ff**and

**Fnet**was solved for :

Fa = 10. Newtons

m = 0.128kg

a = 2.5m/s/s

Ff = ?

Ff was needed later so rather than doing m*a to find Fnet, we found

**Ff**first to find Fnet:

Fnet = ma

Fa + Ff = ma

Ff = 0.128kg(2.5m/s/s)- 10. Newtons

Ff = -9.68 Newtons

**(← rounded to correct amount of sig digs.)**

__= -9.7 Newtons__And now

**Fnet**:

Fnet = Fa-Ff

= 10. Newtons - 9.7 N

__= 0.3 Newtons__At this point, everything seemed reasonable, but then we needed to solve for the coefficient of friction.

## Homework Equations

μk = Fk/Fn

## The Attempt at a Solution

μk = Fk/Fn

= -9.7 Newtons / (0.128)*(9.8)

= 7.73

= 7.7 (← rounded to correct number if sig digs.)

__The question is, does 7.7 make sense for a coefficient of friction, and does everything else look sound?__. The mousetrap car we made seemed to be of low friction , except between the ground and the wheel, where we wanted to maximize friction (balloons were used).

Thanks.

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