A ball with a mass of $40 g$ $40 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $16 \frac{k g}{s^{2}}$ $16 (kg)/s^2$ and was compressed by $\frac{2}{5} m$ $2/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2018-05-25T06:14:37

The height reached by the ball is $= 3.27 m$ $=3.27m$

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The spring constant is $k = 16 k g s^{- 2}$ $k=16kgs^-2$

The compression of the spring is $x = \frac{2}{5} m$ $x=2/5m$

The potential energy stored in the spring is

$P E = \frac{1}{2} k x^{2} = \frac{1}{2} \cdot 16 \cdot {(\frac{2}{5})}^{2} = 1.28 J$ $PE=1/2kx^2=1/2*16*(2/5)^2=1.28J$

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

$K E_{b a l l} = \frac{1}{2} m u^{2}$ $KE_(ball)=1/2m u^2$

Let the height of the ball be $= h$ $=h$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

Then ,

The potential energy of the ball is $P E_{b a l l} = m g h$ $PE_(ball)=mgh$

Mass of the ball is $m = 0.040 k g$ $m=0.040kg$

$P E_{b a l l} = 1.28 = 0.040 \cdot 9.8 \cdot h$ $PE_(ball)=1.28=0.040*9.8*h$

$h = 1.28 \cdot \frac{1}{0.040 \cdot 9.8}$ $h=1.28*1/(0.040*9.8)$

$= 3.27 m$ $=3.27m$

The height reached by the ball is $= 3.27 m$ $=3.27m$

A ball with a mass of 40 g40g40 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16 (kg)/s^216kgs216 (kg)/s^2 and was compressed by 2/5 m25m2/5 m when the ball was released. How high will the ball go?