A child rolls a basketball of mass 0.646 kg up a long ramp. The basketball can b

Question

A child rolls a basketball of mass 0.646 kg up a long ramp. The basketball can be considered a thin-walled hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 7.70 m/s . When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 3.70 m/s . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping.

Part A

Find the maximum vertical height increase of the ball as it rolls up the ramp.

Take free fall acceleration to be g = 9.80 m/s2 .

Explanation / Answer

Kinetic energy of hollow sphere = 0.5*m*v^2 + 0.5*I*w^2

= 0.5*m*v^2 + 0.5*(2/3)*m*r^2*w^2

= 0.5*m*v^2 + (1/3)*m*(r*w)^2

= 0.5*m*v^2 + 0.333*m*v^2

= 0.8333*m*v^2

so, Workdone by friction = change in kinetic energy

= 0.8333*m*(v2^2 - v1^2)

= 0.83333*0.646*(3.7^2 - 7.7^2)

= -24.548 J

so, workdone by fristion on sphere while moving up = -24.548/2

= -12.274 J.

Let Hmax is the maximum height reached.

Apply, Workdone by friction = change in mechaincal energy

-12.274 = m*g*Hmax - 0.83333*m*v1^2

-12.274 = 0.646*9.8*Hmax - 0.8333*0.646*7.7^2

-12.274 = 6.33*Hmax - 0.8333*0.646*7.7^2

-12.274 = 6.33*Hmax - 31.92

Hmax = (31.92 - 12.274)/6.33

= 3.1 m <<<<<<<------------------Answer

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