A toy cannon uses a spring to project a 5.21-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.21-g soft rubber ball. The spring is originally compressed by 5.09 cm and has a force constant of 7.94 N/m. When the cannon is fired, the ball moves 15.3 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 8 N on the ball. With what speed does the projectile leave the barrel of the cannon? m/s At what point does the ball have maximum speed? cm (from its original position) What is this maximum speed? m/s

Explanation / Answer

F = kx = 7.94*0.0509 = 0.404146 N

but F = ma

0.404146 = 0.00521 *a
a = 77.5712 m/s^2

friction force acting :fs = 0.0328N

fs = m ad

0.0328 = 0.00521 *ad

ad =7.2936m/s^2

so net acceleration = a-ad = 77.5712-7.2936 = 70.2776 m/s^2

starts with velocity = 0

s = ut +0.5at2

time take to move out of the barrel

0.153 = 0+0.5*70.2776*t2

t = 0.065986 sec



1) v= u+at

v = 0+70.2776*0.065986 = 4.63733 /sec ---------------------ANSWER1

2) Ball has maximum speed at the end of the barrel

that is 15 cm ----------------------ANSWER2

3) maximum speed = speed at the end of the barrel

maximum speed = 4.63733 m/sec ------------------ANSWER3

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