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 $42 \frac{k g}{s^{2}}$ $42 (kg)/s^2$ and was compressed by $\frac{5}{3} m$ $5/3 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2016-04-12T06:05:45

$h = 21, 25 m$ $h=21,25" m"$

$given data:$ $"given data:"$

$Delta x=5/3 m$

$K=42 (kg)/s^2$

$m=280g=0,28 kg" mass of object"$

$g=9,81 m/s^2$

$h=?$

$E_p=1/2*k*Delta x^2" Potential energy for compressed springs"$

$E=m*g*h " potential energy for object raising from earth"$

$E_p=E" conservation of energy"$

$m*g*h=1/2*K*Delta x^2$

$h=(K*Delta x^2)/(2*m*g)$

$h=(42*(5/3)^2)/(2*0,28*9,81)$

$h=(42*25/9)/(5,49)$

$h=(116,67)/(5,49)$

$h=21,25" m"$

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 42 (kg)/s^242kgs242 (kg)/s^2 and was compressed by 5/3 m53m5/3 m when the ball was released. How high will the ball go?