• stripes
In summary, the conversation discusses finding the work done on a football by a spring loaded cannon, the speed of the football when the spring is at its equilibrium position, and the distance the football will slide before coming to a stop. Different equations, such as F = kx for a spring, W = Fd, Fnet = ma, and v^2 - (v initial)^2 = 2ax are used to solve for these values. The correct method for determining the work done on the football by the spring is by taking the definite integral of the force equation, resulting in 50 joules. The work done by the spring is equal to the change in kinetic energy of the ball, which can also be found using the elastic

## 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...

"W = Fd
W = (.250 N)(.500 m)
W = .125 J

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

The force of the spring is not constant. It is proportional to the compression, so it decreases linearly while the deviation from equilibrium changes from 0.50 meter to zero. How do you get the work? Moreover, the spring constant is not 500 N/m.

ehild

lol, i just realized that k = 400 N/m, and (400 N/m)(.500 m) is 100 N...though this isn't really important. We see that because force is not constant, we must take the area under the curve, that is, the definite integral, of this force equation Fspring = 400x on the interval 0 to .500 m. Doing so gives you 200x^2 from 0 to .500, which gives you 50 joules...so the work done over that change in x is 50 joules, correct?

From there, is it safe to say I can use the same methods I did above to solve for (b) and (c)?

Of course, you can apply this method for question b. You know the work done on the ball; this work will be equal to the change of the kinetic energy according to the work-energy theorem.

Or you can think in terms of energy. The spring did work on the ball on the account of its elastic energy - but it is equal to the work which was needed to compress the spring. Some times before, the spring was compressed, work of 1/2 k x^2 was done and that work gave the spring so much energy.When the spring is released, this elastic energy transforms into other kind of energy, for example, the kinetic energy of the ball. You can use this formula E(spring)= 1/2 kx^2 whenever a compressed spring is concerned.

You determine v from the kinetic energy of the ball. With this initial velocity, the ball moves with constant deceleration under the effect of friction. When will it stop?

ehild

ahhh...kinetic energy of a spring...should have thought of that, would have saved me a lot of trouble. Either way, the question has been handed in, and we'll see how it turns out!

Thanks so much ehild for the help!

stripes said:
ahhh...kinetic energy of a spring...should have thought of that, would have saved me a lot of trouble. Either way, the question has been handed in, and we'll see how it turns out!

Thanks so much ehild for the help!

it is elastic energy of the spring, not kinetic. 