Popgun KE & PE Homework: Consider the Figure

[illidari]

Homework Statement

Consider the popgun in the figure below. The launching mechanism consists of a trigger-released spring. The spring is compressed to a position yA, and the trigger is fired. The projectile of mass m rises to a position yC above the position at which it leaves the spring, indicated in the figure as position yB = 0. Consider a firing of the gun for which m = 38.0 g, yA = -0.140 m, and yC = 22.0 m. Suppose, however, there is a friction force of magnitude 1.90 N acting on the projectile as it rubs against the interior of the barrel. The vertical length from point to the end of the barrel is 0.620 m. (Hint: you may need to calculate the spring constant from the information given.)


(a) After the spring is compressed and the popgun fired, to what height does the projectile rise above point ?


http://www.webassign.net/serpse8/8-figure-06-alt.gif

Picture

Homework Equations

Book gave this for no friction to solve for K
K(c)+Ug(c)+Us(c)=K(A)+Ug(A)+Us(A)

0 + mgy(c) + 0 = 0 + mgy(a) + 1/2kx^2

k=2mg(y(c)-y(a))/x^2

The Attempt at a Solution

I am confused how to use the above with friction involed. It's saying at Point C, KE = 0? Wouldn't that mean it reached it's maximum height ? (Assume this is wrong, would be to easy then)

Do I do :

Change in KE + Change in PE - Ffr* D for the length of the gun to find the velocity it leaves barrel and then use the velocity find how high that one goes ?

 

[PhanthomJay]

Homework Statement

Consider the popgun in the figure below. The launching mechanism consists of a trigger-released spring. The spring is compressed to a position yA, and the trigger is fired. The projectile of mass m rises to a position yC above the position at which it leaves the spring, indicated in the figure as position yB = 0. Consider a firing of the gun for which m = 38.0 g, yA = -0.140 m, and yC = 22.0 m. Suppose, however, there is a friction force of magnitude 1.90 N acting on the projectile as it rubs against the interior of the barrel. The vertical length from point to the end of the barrel is 0.620 m. (Hint: you may need to calculate the spring constant from the information given.)


(a) After the spring is compressed and the popgun fired, to what height does the projectile rise above point ?

Homework Equations

Book gave this for no friction to solve for K
K(c)+Ug(c)+Us(c)=K(A)+Ug(A)+Us(A)

0 + mgy(c) + 0 = 0 + mgy(a) + 1/2kx^2

k=2mg(y(c)-y(a))/x^2

There is an error in sign in one of these terms..did you copy it correctly and understand why?

The Attempt at a Solution

I am confused how to use the above with friction involed. It's saying at Point C, KE = 0? Wouldn't that mean it reached it's maximum height ?

Yes, Point C is the point of max height.

Do I do :

Change in KE + Change in PE - Ffr* D = 0[/color] for the length of the gun to find the velocity it leaves barrel and then use the velocity find how high that one goes ?

You don't need to find the velocity as it leaves the barrel. Just apply your equation between points A and C to solve for the max height. Please watch plus/minus signs.

 

[illidari]

I copied it correctly from what I could tell. It took gpe to be 0 when it leaves the spring. Any chance you could help me out in pointing out where the sign error is?

The question is asking for the max height it goes above point y(b)and point y(c) is given.
y(b)=0 , y(c)=22

22-0 = 22m <- answer?

It doesn't feel like this problem would be a simple read and copy a value given to you :(?

Is the 22m the max height without friction and I am solving for what friction removed from it?

 

[PhanthomJay]

I copied it correctly from what I could tell. It took gpe to be 0 when it leaves the spring. Any chance you could help me out in pointing out where the sign error is?

If you take the gpe as 0 when the projectile leaves the spring at B, then the gpe of the projectile in its starting position 0.14 m below B must be negative. Maybe you have that right...plug in the numbers to solve for k, the spring contant, and I'll check your plus and minus sign.

The question is asking for the max height it goes above point y(b)and point y(c) is given.
y(b)=0 , y(c)=22

22-0 = 22m <- answer?

It doesn't feel like this problem would be a simple read and copy a value given to you :(?

There are 2 parts to the problem. In the first part, it is given that y(c) = max height =22 m, and you solve for k. Then in the 2nd part, you include friction, you have k, now solve for y(c), the max height when you include barrel friction. It must be less than 22 m.

 

[illidari]

If you take the gpe as 0 when the projectile leaves the spring at B, then the gpe of the projectile in its starting position 0.14 m below B must be negative. Maybe you have that right...plug in the numbers to solve for k, the spring contant, and I'll check your plus and minus sign.
There are 2 parts to the problem. In the first part, it is given that y(c) = max height =22 m, and you solve for k. Then in the 2nd part, you include friction, you have k, now solve for y(c), the max height when you include barrel friction. It must be less than 22 m.

k = 2(.038kg)*9.8*[22-(-.14)] / .14^2 = 841.32 N/m

Book example has different values for their problem and they got 958 N/m Looks like this is near enough.

So

K(C)+Ug(C)+Us(C)= K(A)+Ug(A)+Us(A)

0+mgy(C) + 0 = 0 + mgy(A) + 1/2kx^2

mgy(C)= mgy(A) + 1/2 * 841.32* .14^2

mgy(c)= mgy(A) + 8.244936

mgy(C) - mgy(A) -8.244936= -Ffr * d

.038 * 9.8 * y(C) - .038 * 9.8 * (-.14) - 8.244936 + (1.9 *.620) =0

.3724 * y(c) + .052136 - 8.244936 + 1.178 = 0

.3724 *y(C) = 7.0148

y(c)= 18.8 =) It took this answer.

Thanks dude , you rule for helping me out on that.

 

[PhanthomJay]

k = 2(.038kg)*9.8*[22-(-.14)] / .14^2 = 841.32 N/m

Book example has different values for their problem and they got 958 N/m Looks like this is near enough.

They apparently used g=10m/s^2, and solved for k = 858 N/m, close enough. You successfully handled the minus sign...

So

K(C)+Ug(C)+Us(C)= K(A)+Ug(A)+Us(A) -Ffr * d[/color]

0+mgy(C) + 0 = 0 + mgy(A) + 1/2kx^2 -Ffr * d[/color]

mgy(C)= mgy(A) + 1/2 * 841.32* .14^2 -Ffr * d[/color]

mgy(c)= mgy(A) + 8.244936 -Ffr * d[/color]

mgy(C) - mgy(A) -8.244936= -Ffr * d

.038 * 9.8 * y(C) - .038 * 9.8 * (-.14) - 8.244936 + (1.9 *.620) =0

.3724 * y(c) + .052136 - 8.244936 + 1.178 = 0

.3724 *y(C) = 7.0148

y(c)= 18.8 =) It took this answer.

Thanks dude , you rule for helping me out on that.

Yes, nice work...Make sure that Ffr * d term carries through all steps, as i noted in red...you left it out in the initial steps, then put it in at the end...but in the end, it came out OK...