A ball of mass .8kg is pressed against a compressed spring and then the ball and

Question

A ball of mass .8kg is pressed against a compressed spring and then the ball and spring are released from rest. The spring apples a constant force of .26N over a distance of .3meters. The ball then proceeds to roll along a frictionless surface until it encounters a ramp which rises 26 degrees about the horizontal and has a coefficient of rolling friction of .06. 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? Please show steps

Explanation / Answer

A) The acceleration of the ball = a = F/m = 0.26/0.8 = 0.325 m/s^2

B) The speed of the ball = v =sqrt(2ad) = sqrt(2*0.325*0.3) = 0.442 m/s

C) 0.5mv^2 + 0.5 Iw^2 = mgh + umgh/sin(theta)

Now I = 0.4mR^2 and w = v/R

Then 0.5Iw^2 =0.2 mv^2

So, 0.5mv^2 + 0.2 mv^2 = mgh + umgh/sin(theta)

0.7*0.8*0.442^2 = 0.8*9.8*h[1 + 0.06/sin26]

0.10940384 = 8.91*h

h = 0.0123 m = 1.23 cm

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