A ball with a mass of #80 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #9 (kg)/s^2# and was compressed by #3/5 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #80 g# is projected vertically by a spring loaded contraption.  The spring in the contraption has a spring constant of #9  (kg)/s^2# and was compressed by #3/5 m# when the ball was released.  How high will the ball go?

Amelia Adams · Accepted Answer

#h=81/8m#

Given that the entire spring energy is converted to kinetic energy which later turns to potential energy. So that means the entire spring energy is turns to potential energy To solve, we'll use the near-earth potential equation.

Now, spring energy equation is #E=1/2kx^2# We have, #k=9kg/s^2#, #x=3/5m#. So #E=1/2*9*(3/5)^2# So, we end up with #E=3^4/10J#

Now, The near-earth potential energy is #E=mgh# #m=80g=80*10^{-3}kg#, #g=10m/s^2#, so we need to find #h#

So, equating the two, we get #3^4/cancel10^1= cancel80^8*cancel10^1*cancel10^{-3}*h#

Re-arranging gives us the answer that I have up there. You'll need a calculator to find it in decimals.

Adrian Ayala · Answer

undefined To find the maximum height the ball will reach, you can use the principle of conservation of mechanical energy. The potential energy stored in the compressed spring is converted into kinetic energy as the ball is released and moves upward. At the maximum height, all of the kinetic energy is converted back into potential energy.

The potential energy stored in the compressed spring is given by:

$PE_{spring} = \frac{1}{2} kx^2$

where $k$ is the spring constant and $x$ is the compression distance.

Given $k = 9 \, \text{(kg/s}^2$) and $x = \frac{3}{5} \, \text{m}$, we can calculate the potential energy stored in the spring.

$PE_{spring} = \frac{1}{2} \times 9 \times \left(\frac{3}{5}\right)^2$

Now, we can use the conservation of mechanical energy to find the maximum height the ball will reach. The total mechanical energy (kinetic energy plus potential energy) remains constant throughout the motion. At the maximum height, all of the kinetic energy is converted into potential energy.

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

Since the ball is initially at rest when released, its initial kinetic energy is zero.

$PE_{spring} = PE_{final}$

$PE_{final} = mgh$

Where $m$ is the mass of the ball, $g$ is the acceleration due to gravity, and $h$ is the maximum height reached by the ball.

Given $m = 80 \, \text{g}$ ($0.08 \, \text{kg}$) and $g = 9.8 \, \text{m/s}^2$, we can solve for $h$.

$PE_{spring} = mgh$

$h = \frac{PE_{spring}}{mg}$

Substitute the values:

$h = \frac{\frac{1}{2} \times 9 \times \left(\frac{3}{5}\right)^2}{0.08 \times 9.8}$

$h \approx 0.347 \, \text{m}$

So, the ball will reach a maximum height of approximately $0.347 \, \text{m}$.