A toy cannon uses a spring to project a 5.22-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.22-g soft rubber ball. The spring is originally compressed by 5.07 cm and has a force constant of 8.10 N/m. When the cannon is fired, the ball moves 14.8 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 5 N on the ball (a) With what speed does the projectile leave the barrel of the cannon? 1.41 Make certain you have correctly converted all of the centimeter and gram quantities to MKS units. m/s (b) At what point does the ball have maximum speed? cm (from its original position) (c) What is this maximum speed? Need Help?Read It

Explanation / Answer

(a) m = 5.22 g = 0.00522 kg

First calculate the force applied by the spring on the ball.

This is -

Fs = k x
=>Fs = 8.10 x 5.07/100 = 0.41067 N

Net Force on ball (Fn) = Fs - F(friction)

=>Fn = 0.41067 - 0.0325 = 0.37817 N

again we have -

Fn = m x a
=>0.37817 = 0.00522 x a
=>a = 0.37817 / 0.00522 = 72.45 m/s^2

use the expression -

v^2 = u^2 + 2as
=>v^2 = 0 + 2 x 72.45 x 14.8 /100
=>v^2 = 10.72
=>v = 3.27 m/s

so the speed of the projectile when it leaves the barrel = 3.27 m/s.

(b) The max. velocity of ball will be at the point of leaving the barrel.
(c) Maximum speed = 3.27 m/s.

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