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The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1300 N/m that he will compress with a force of 6500 N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 50 N during the 5.0 mm he moves in the barrel.

Required:
At what speed will he emerge from the end of the barrel, 2.5 mabove his initial rest position?

User Dedoki
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1 Answer

3 votes

Answer:

22m/s

Step-by-step explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0


F_x=kx

Substitute the values


6500=1300x


x=(6500)/(1300)=5m

Work done due to friction force


W_f=fscos\theta

We have
\theta=180^(\circ)

Substitute the values


W_f=50* 5cos180^(\circ)


W_f=-250J

Initial kinetic energy, Ki=0

Initial gravitational energy,
U_(grav,1)=0\

Initial elastic potential energy


U_(el,1)=(1)/(2)kx^2=(1)/(2)(1300)(5^2)


U_(el,1)=16250J

Final elastic energy,
U_(el,2)=0

Final kinetic energy,
K_f=(1)/(2)(60)v^2=30v^2

Final gravitational energy,
U_(grav,2)=mgh=60* 9.8* 2.5

Final gravitational energy,
U_(grav,2)=1470J

Using work-energy theorem


K_i+U_(grav,1)+U_(el,1)+W_f=K_f+U_(grav,2)+U_(el,2)

Substitute the values


0+0+16250-250=30v^2+1470+0


16000-1470=30v^2


14530=30v^2


v^2=(14530)/(30)


v=\sqrt{(14530)/(30)}


v=22m/s

User Emachine
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