In a spring gun, a spring of force constant 430 N/m is compressed 0.10 m . When

Question

In a spring gun, a spring of force constant 430 N/m is compressed 0.10 m . When fired, 79.5 %of the elastic potential energy stored in the spring is eventually converted into kinetic energy of a 6.20×102 kg uniform ball that is rolling without slipping at the base of a ramp. The ball continues to roll without slipping up the ramp with 89.9 % of the kinetic energy at the bottom converted into an increase in gravitational potential energy at the instant it stops. (please no copy and pasting, and diagram would be helpful to understand whats going on in the problem)

Part A

What is the speed of the ball's center of mass at the base of the ramp?

Part B

At this position, what is the speed of a point at the top of the ball?

Part C

At this position, what is the speed of a point at the bottom of the ball?

Part D

What maximum vertical height up the ramp does the ball move?

Explanation / Answer

a)

kinetic energy at the base is given as

KE = (0.5) m v2 + (0.5) I w2 = (0.5) m v2 + (0.5) (2/5) (mr2) (v/r)2

KE = (0.7) m v2

spring potential energy is given as

Es = (0.5) k x2

Given that : KE = (0.795) Es

(0.7) m v2 = (0.795) (0.5) k x2

(0.7) (0.062) v2 = (0.795) (0.5) (430) (0.1)2

v = 6.3 m/s

b)

at the Top

Vt = 2 v = 2 x 6.3 = 12.6 m/s

c)

at the bottom :

Vb = 0

d)

using conservation of energy

Potentoal energy = PE = 0.899 KE

PE = (0.899) ((0.795) Es)

mgh = (0.715) (0.5) (430) (0.1)2

(0.062) (9.8) h = 1.54

h = 2.53 m

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