A ball with a mass of $140 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $21 (kg)/s^2$ and was compressed by $3/8 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-10-21T03:12:58

1.076 m

Potential stored in spring $= (kx^2)/2 = (21 kg.s^-2 × (3/8 m)^2)/2 = 189/128J$

Kinetic energy gained by ball $=(mv^2)/2$
$= (140 cancel(g) × 10^-3 (kg)/cancel(g) × v^2)/2 = (0.07 kg) × v^2$

According to law of conservation of energy
Potential energy in spring = Kinetic energy gained by ball

$189/128 J = (0.07 kg) × v^2$

$v^2 = 21.09375 J/(kg)$

$H = v^2/(2g) = (21.09375 J/(kg)) / (2 × 9.8 ms^-2)= 1.076 m$

A ball with a mass of 140 g140 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 21 (kg)/s^221 (kg)/s^2 and was compressed by 3/8 m3/8 m when the ball was released. How high will the ball go?