# Physics mechanics Question.?

A toy cannon uses a spring to project a 5.27 g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force constant of 7.90 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and there is a constant frictional force of 0.0317 N between the barrel and the ball.

A.) With what speed does the projectile leave the barrel of the cannon?

B.) At what point does the ball have maximum speed? in cm from its original position

C.)What is this maximum speed?

I tried getting the work done by the spring and subtracting the work done by the frictional force. then i set that equal to 1/2mv^2 . and i was wrong. i don't know what other path to take.

### 1 Answer

- JimLv 74 years agoFavorite Answer
So cannon is fired HORIZONTALLY. :>)

SPE = 1/2kx² = (0.5)(7.90)(0.0501)² = 9.91454E-3 J

Work done against friction = (0.0317)(0.150) = 4.755E-3 J

KE of soft rubber ball as it leaves barrel of cannon = (9.91454 - 4.755)E-3 = 5.160E-3 J

ball's speed leaving barrel = V = √2KE/m = √[10.3191E-3/0.00527] = √1958.08E-3 ≈ 1.40 m/s ANS A.)

ball has max speed as it leaves the spring...ie after moving 5.01 cm from its initial in barrel position ANS B.)

Vmax = √2(SPE)/m = √ 3762.634 ≈ 61.3 m/s ANS C.)

Thanks! i forgot to square the x value in my spring potential energy and that screwed up my whole answer.