A ball with a mass of $280 g$ $280 g$ 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 $\frac{7}{3} m$ $7/3 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2018-04-03T06:19:43

The height is $= 15.87 m$ $=15.87m$

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

The compression is $x = \frac{7}{3} m$ $x=7/3m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 16 \cdot {(\frac{7}{3})}^{2} = 43.56 J$ $PE=1/2*16*(7/3)^2=43.56J$

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 $m = 0.280 k g$ $m=0.280kg$

$P E_{b a l l} = 43.56 = 0.28 \cdot 9.8 \cdot h$ $PE_(ball)=43.56=0.28*9.8*h$

$h = 43.56 \cdot \frac{1}{0 . .28 \cdot 9.8}$ $h=43.56*1/(0..28*9.8)$

$= 15.87 m$ $=15.87m$

The height is $= 15.87 m$ $=15.87m$

A ball with a mass of 280 g280g280 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 7/3 m73m7/3 m when the ball was released. How high will the ball go?