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## Homework Statement

A football with a mass of 0.40 kg is placed into a spring loaded cannon with k = 400 N/m. The spring is compressed by 50.0 cm. a.) find the work done on the football by the spring as the football moves from the compressed position to its equilibrium position. The cannon is aimed horizontally along the icy grass of a football field. b.) Find the speed of the football when the spring is at its equilibrium position. The cannon is aimed horizontally along the icy grass of the football field. If the coefficient of kinetic friction between the football and the icy grass is 0.080, c.) how far down the field will the football slide before coming to a stop?

## Homework Equations

F = kx (for a spring)

W = Fd

Fnet = ma

v^2 - (v initial)^2 = 2ax

## The Attempt at a Solution

a.) F = kx (for a spring)

F = (500 N/m)(.500 m)

F = .250 N

W = Fd

W = (.250 N)(.500 m)

W = .125 J

This was marked wrong...where did I make my mistake?

b.) Fnet = ma

Fspring = ma

a = Fspring/m

a = (.250 N)/(.40 kg)

a = .625 m/s^2

v^2 - (v initial)^2 = 2ax

v = sqrt(2ax + (v initial)^2)

v = sqrt(2(.625 m/s^2)(.500 m) + (0 m/s))

v = .79 m/s

This was wrong...

c.) Fnet = ma

Fspring - uFn = ma <<<

**uFn is force of friction, u is coefficient of kinetic friction, Fn is normal force.**

Fspring - u(mg) = ma

a = (Fspring - u(mg))/m

a = (.250 N - (.080)(.40 kg)(9.81 m/s^2))/(.40 kg)

a = -.1598 m/s^2 <<< so the football is decelerating as it goes down the icy football field.

v^2 - (v initial)^2 = 2ax

x = (v^2 - (v initial)^2)/2a

x = (0^2 - (.79 m/s)^2)/2(-.1598 m/s^2) <<< info from above and part (b)

x = 1.95 meters

This was also marked wrong...

I'm not quite understanding why this method of solving is incorrect...

Thanks so much for your help in advance!