A ball of mass 0.7 kg is pressed against a compressed spring and then the ball a

Question

A ball of mass 0.7 kg is pressed against a compressed spring and then the ball and spring are released from rest. The spring applies a constant force of 0.25 Newtons over a distance of 0.3 meters. The ball then proceeds to roll along a frictionless surface until it encounters a ramp which rises 25 degrees above the horizontal and has a coefficient of rolling friction of r = 0.05. (A) What is the acceleration of the ball while the spring is pushing it? (B) What is the speed of the ball immediately before it rolls onto the ramp? (C) How high above the horizontal does the ball roll?

Explanation / Answer

The work produced by the spring will be

W = Fd
W = 0.25(0.3)
W = 0.075 J

which will result in a change of kinetic energy

KE = ½mv²
v² = 2KE/m

v² = 2(0.075)/0.7
v² = 0.214 m2/s²

using a kinematic equation
v² = v² + 2as
a = v² / 2s

a = 0.214 / 2(0.3)
a = 0.357 m/s²

round to the appropriate precision

OR

use something basic like
F = ma
a = F/m

a = 0.25 N / 0.7 kg
a = 0.357 m/s²

b) the speed of the ball immediately before it rolls onto the ramp, v=sqrt(0.214)=0.4626 m/s

c)assume the ball can roll upto h height

0.075= mgh+f*(h/sin25) (where f=friction force = umgcos25)

0.075=h(0.7*9.8+0.05*0.7*9.8*cot25)

h=9.8*10^-3 m=0.98 cm

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