A spring with spring constant 24 N/m is compressed a distance of 6.5 cm by a bal

Question

A spring with spring constant 24 N/m is compressed a distance of 6.5 cm by a ball with a mass of 213.5 g (see figure below). The ball is then released and rolls without slipping along a horizontal surface, leaving the spring at point A. The process is repeated, using a block instead, with a mass identical to that of the ball. The block compresses the spring by 6.5 cm and is also released, leaving the spring at point A. Assume the ball rolls, but ignore other effects of friction. (Assume that the ball is a solid ball.) (a) What is the speed of the ball at point B? m/s (b) What is the speed of the block at point B? m/s

Explanation / Answer

a)

Using law of conservation of energy

energy stored in spring = kinetic energy of ball

0.5*k*x^2 = Rotational kinetic energy + trnaslational kinetic energy

0.5*k*x^2 = (0.5*I*w^2)+(0.5*m*v^2)

I is the moment of inertia = (2/5)*m*r^2 = (2/5)*213.5*10^-3*r^2

w = v/r

then

0.5*k*x^2 = (0.5*(2/5)*213.5*10^-3*r^2*(v/r)^2)+(0.5*m*v^2)

0.5*24*0.065^2 = (0.5*(2/5)*213.5*10^-3*v^2)+(0.5*213.5*10^-3*v^2)

v = 0.582 m/sec is the answer for A)

b)

incase of block

energy stored in spring = kinetic energy of block

0.5*k*x^2 = 0.5*m*v^2

0.5*24*0.065^2 = 0.5*213.5*10^-3*v^2

v = 0.689 m/sec

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