A ball with a mass of #300 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 #7/6 m# when the ball was released. How high will the ball go?

Question

A ball with a mass of #300 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 #7/6 m# when the ball was released.  How high will the ball go?

Nathan Ashcroft · Accepted Answer

Maximum altitude #~~2.8m#

Ignoring air resistance, other sources of friction, etc., this problem can be solved with energy conservation:

#E_(mech)=K+U#

Where #K# is kinetic energy and #U# is potential energy. Thus, when energy is conserved in a system it should follow that

#E_i=E_f#

#=>K_i+U_i=K_f+U_f#

From this relationship we can see that if potential energy, for example, were to decrease, kinetic energy would have to increase in order for energy to be conserved. The inverse is also true. The idea is that energy is being transformed between two forms (potential and kinetic) rather than any of it being lost of gained. When we have to account for friction, air resistance, etc., energy is not conserved, because some of it is lost to heat (#E_(th)#). In reality, energy is "lost" to a variety of other sources as well, such as in the production of sound.

Initially, assuming the spring is compressed to point where its height is insignificant (so that we need not worry about gravitational potential energy of the ball beforehand), all energy is stored as spring potential energy in the spring. Because nothing is moving prior to the launch, there is no kinetic energy. Therefore, we have only potential energy to begin with.

#U_(sp)=U_f+K_f#

After the launch, as a the spring decompresses, the energy stored within it as spring potential energy is transferred to the ball, which is observed as the ball shoots upward. The spring potential energy has been transformed into kinetic energy. However, as the ball rises, it is gaining altitude, and therefore gaining gravitational potential energy. Because the potential energy of the ball increases, its kinetic energy must decrease, and we observe this in the form of the ball stopping at some point in the air before falling back to the earth. This is the point at which all of the kinetic energy has been transformed into gravitational potential energy.

Because we are concerned with the maximum altitude of the ball, what we're really asking is *at what point all of the potential energy from the spring has been transferred into gravitational potential energy, * which is given by #U_g=mgh#. When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy (remember projectile motion?), and therefore we ultimately have only gravitational potential energy. Our statement of energy conservation is then

#U_(sp)=U_g#

Spring potential energy, #U_(sp)# is given by #1/2k(Δs)^2#, where #k# is the spring constant of the spring and #Deltas# is its displacement from equilibrium (length).

#1/2k(Δs)^2=mgh#

( tl;dr ) Rearranging to solve for #h#,

#=>h=(1/2k(Δs)^2)/(mg)#

Using our known values for mass, the spring constant, and the compression of the spring, we can now calculate a value for #h#.

#h=(1/2(12N/m)(7/6m)^2)/(0.300kg*9.81m/s^2)#

#~~2.8m#

Charles Aeschliman · 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 spring when compressed is converted into kinetic energy of the ball when released, and then into gravitational potential energy at the maximum height.

The potential energy stored in the compressed spring is given by $ \frac{1}{2} kx^2 $, where $ k $ is the spring constant and $ x $ is the compression distance.

Given:
- Mass of the ball ($ m $): 300 g = 0.3 kg
- Spring constant ($ k $): 12 (kg/s^2)
- Compression distance ($ x $): $ \frac{7}{6} $ m

The potential energy stored in the spring is:
$$ \text{Potential energy} = \frac{1}{2} kx^2 $$

Then, equate this to the gravitational potential energy at the maximum height:
$$ mgh = \frac{1}{2} kx^2 $$

Solve for $ h $, which represents the maximum height the ball reaches.

Adrian Ayala · Answer

undefined To find the maximum height the ball will reach, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is equal to the final kinetic energy of the ball at its maximum height.

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

$$ PE = \frac{1}{2}kx^2 $$

where $ k $ is the spring constant and $ x $ is the compression of the spring. Plugging in the values:

$$ PE = \frac{1}{2} \times 12 \times \left( \frac{7}{6} \right)^2 $$

$$ PE = \frac{1}{2} \times 12 \times \frac{49}{36} $$

$$ PE = 7 \, \text{J} $$

At maximum height, all the initial kinetic energy of the ball is converted into potential energy. The initial kinetic energy of the ball is given by:

$$ KE = \frac{1}{2}mv^2 $$

where $ m $ is the mass of the ball and $ v $ is its velocity at the release point. Since the ball is projected vertically, its initial velocity is 0.

Therefore, the initial kinetic energy is also 0.

At the maximum height, the potential energy equals the initial potential energy:

$$ PE = mgh $$

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

Setting $ PE $ equal to $ mgh $ and solving for $ h $:

$$ mgh = 7 $$

$$ 0.3 \times 9.8 \times h = 7 $$

$$ h = \frac{7}{0.3 \times 9.8} $$

$$ h ≈ 2.38 \, \text{m} $$

So, the ball will reach a maximum height of approximately 2.38 meters.