In a physics lab experiment, a compressed spring launches

[Bigworldjust]

Homework Statement

In a physics lab experiment, a compressed spring launches a 40g metal ball at a 25 degree angle. Compressing the spring 17cm causes the ball to hit the floor 1.6m below the point at which it leaves the spring after traveling 5.3m horizontally.

What is the spring constant?
Express your answer using two significant figures.

Homework Equations

y = x tan Φ + ½ g (x/(Vo cos Φ))²
½ m Vo² - ½ k (Δx)² = 0

The Attempt at a Solution

x = Vo t cos Φ

t = x/(Vo cos Φ)

y = Vo t sin Φ + ½ g t²

y = Vo(x/(Vo cos Φ)) sin Φ + ½ g (x/(Vo cos Φ))²

y = x tan Φ + ½ g (x/(Vo cos Φ))²

-1.6 = 5.3 tan 25° + ½ (-9.8)(5.3/Vo)² sec² 25°

Vo = 6.41541 m/s

according to conservation of energy,

ΔE = 0

½ m Vo² - ½ k (Δx)² = 0

½ (0.040)(6.41541)² - ½ k (0.17)² = 0

k = 56.9654 N/m

Since the question asked for two significant figures I put in 57, and I got it wrong with an explanation saying that it was close to the answer but I may have made an error in significant figures or approximation. Could someone please tell me where my slight error is? Thank you!

 

[Bigworldjust]

Bump, any help =/?

 

[Bigworldjust]

I have to hand the assignment in soon, so any urgent help would be appreciated, lol.

 

[gneill]

Spring potential energy isn't the only form of potential energy doing its thing here. Gravity is working against the spring's efforts...

 

[Bigworldjust]

Spring potential energy isn't the only form of potential energy doing its thing here. Gravity is working against the spring's efforts...

So how would I go about adding that in?

 

[gneill]

So how would I go about adding that in?

How much gravitational PE is involved during the launch while the ball is in the launcher? Does that PE get added to the ball's KE or stolen from it?

 

[Bigworldjust]

How much gravitational PE is involved during the launch while the ball is in the launcher? Does that PE get added to the ball's KE or stolen from it?

Hmm, I'm not sure, lol. Wouldn't the PE be converted to KE?

 

[gneill]

When you throw a ball into the air, does change in gravitational PE add to or subtract from the ball's speed? (Does it go faster as it rises, or does it slow down?)

 

[Bigworldjust]

When you throw a ball into the air, does change in gravitational PE add to or subtract from the ball's speed? (Does it go faster as it rises, or does it slow down?)

It would be slowing down as it rises so you would subtract the PE from the ball's speed?

 

[gneill]

It would be slowing down as it rises so you would subtract the PE from the ball's speed?

Yes. (Technically, the ball's KE is being converted into gravitational PE as it rises in the gravitational field).

 

[Bigworldjust]

Yes. (Technically, the ball's KE is being converted into gravitational PE as it rises in the gravitational field).

Ah okay, how would I calculate the PE in this case?

 

[gneill]

Ah okay, how would I calculate the PE in this case?

The change in gravitational PE is m*g*Δh

 

[Bigworldjust]

The change in gravitational PE is m*g*Δh

Alright so that would be .04*9.8*1.6?

 

[gneill]

Alright so that would be .04*9.8*1.6?

No, you're only interested in the PE associated with the launch. You've determined what KE the ball should have when it comes out of the launcher, so you're done with the all the rest of the trajectory. You have to concentrate on how to get the launcher to produce the required KE in the ball.

Think of it this way, The PE stored in the spring is going to end up in two places. One is the (desired) KE of the ball so that it can carry out its required trajectory, and the other is the increase in PE the ball obtains during the launch process (while it is being launched).

 

[Bigworldjust]

No, you're only interested in the PE associated with the launch. You've determined what KE the ball should have when it comes out of the launcher, so you're done with the all the rest of the trajectory. You have to concentrate on how to get the launcher to produce the required KE in the ball.

Think of it this way, The PE stored in the spring is going to end up in two places. One is the (desired) KE of the ball so that it can carry out its required trajectory, and the other is the increase in PE the ball obtains during the launch process (while it is being launched).

So you don't use 1.6 as the change in height? Would you use 17cm because of compressing the spring?

 

[gneill]

So you don't use 1.6 as the change in height? Would you use 17cm because of compressing the spring?

That's right. And you don't use all of the 17cm either... the spring is at a 25° angle, so the change in height of the ball will not be the full 17cm.

 

[Bigworldjust]

That's right. And you don't use all of the 17cm either... the spring is at a 25° angle, so the change in height of the ball will not be the full 17cm.

Alright so I would do .17cos(25) to get .154 plug that into m*g*h to get .060368 and subtract that from Vo to get 6.355042 and plug that into get:

½ (0.040)(6.355042)² - ½ k (0.17)² = 0

And than finally to arrive at k to be 56 N/m instead?

 

[gneill]

Alright so I would do .17cos(25) to get .154 plug that into m*g*h to get .060368 and subtract that from Vo to get 6.355042 and plug that into get:

½ (0.040)(6.355042)² - ½ k (0.17)² = 0

And than finally to arrive at k to be 56 N/m instead?

Cosine gives horizontal displacement, not vertical. Check your sketch (you did make a sketch, right? ) You want to find out how much energy gets "lost" to PE when the ball changes height within the launcher. You still want the ball to have the correct KE when it leaves the launcher so that its "initial" velocity will be as you calculated. That is necessary if the original trajectory is to be followed. So leave Vo and KE alone. Deal with adjusting the spring PE to allow for the loss to gravitational PE.

mgh yields an energy (Joules), not a velocity. So you can't just add or subtract it from velocity -- you can add or subtract it from KE's or PE's though