Question

The Great Sandini is a 60.0-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 1100 N/m that he will compress with a force of 4400 N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40.0 N during the 4.00 m he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.50 m above his initial rest position?

Answer 1

Answer:

15.5 m/s

Explanation:

The final kinetic energy (K) of the man will be equal to the initial elastic potential energy stored in the spring (Ue) minus the work done by friction (W) minus the gravitational potential energy lost (Ug), since the man comes out at a position higher than the initial one. Mathematically:

$K=U_e - W - U_g$

Let's now write the various terms:

- Kinetic energy: $K=\frac{1}{2}mv^2$, where m = 60.0 kg is the mass of the man and v is the final speed

- Elastic potential energy: $U_e = \frac{1}{2}kx^2$, where k = 1100 N/m is the spring constant and x the compression of the spring. The compression is given by Hooke's law:

$x=\frac{F}{k}$

And substituting the force applied, F = 4400 N, we find

$x=\frac{4400}{1100}=4.0 m$

So the elastic potential energy is

$U_e=\frac{1}{2}(1100)(4.0)^2=8800 J$

- Work done by friction: $W=F_f d = (40.0)(4.0)=160 J$, where F = 40.0 N is the force of friction and d = 4.0 m is the distance covered

- The change in gravitational potential energy is $U_g = mgh = (60.0)(9.8)(2.50)=1470 J$, where h = 2.50 m is the change in height

So now we can find the kinetic energy of the man:

$K=8800-160-1470=7170 J$

And using the formula for the kinetic energy, we find the speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(7170)}{60}}=15.5 m/s$