A ball with a mass of #450 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #12 (kg)/s^2# and was compressed by #4/3 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #450 g# is projected vertically by a spring loaded contraption.  The spring in the contraption has a spring constant of #12  (kg)/s^2# and was compressed by #4/3 m# when the ball was released.  How high will the ball go?

Charlotte Aaron · Accepted Answer

The height is #=2.42m#

The spring constant is #k=12kgs^-2#

The compression is #x=4/3m#

The potential energy is

#PE=1/2*12*(4/3)^2=32/3J#

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

#KE=1/2m u^2#

The initial velocity is #=u#

#u^2=2/m*KE=2/m*PE#

#u^2=2/0.45*32/3=47.41#

#u=sqrt47.41=6.89ms^-1#

Resolving in the vertical direction #uarr^+#

We apply the equation of motion

#v^2=u^2+2ah#

At the greatest height, #v=0#

and #a=-g#

So,

#0=47.41-2*9.8*h#

#h=47.41/(2*9.8)=2.42m#

Charles Aeschliman · Answer

undefined Use the conservation of energy principle. Calculate potential energy stored in the compressed spring, then equate it to gravitational potential energy at the maximum height. Solve for height (h).

$$ h = \frac{k \cdot x^2}{2 \cdot g \cdot m} $$

Where:
$ k $ = spring constant (12 (kg)/s²)
$ x $ = compression of the spring (4/3 m)
$ g $ = acceleration due to gravity (9.8 m/s²)
$ m $ = mass of the ball (0.45 kg)

$$ h = \frac{12 \cdot (4/3)^2}{2 \cdot 9.8 \cdot 0.45} $$

$$ h \approx 1.78 \, \text{m} $$