A ball with a mass of #150 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 #150 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?

Layla Ahmed · Accepted Answer

The maximum altitude of the ball is #~~1.1m# above the launch point.

Ignoring air resistance, friction, etc., this problem can be solved with energy conservation. I will use the explanation I gave for a similar question in the past. Find tl;dr to skip mini lesson in 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 include friction in problems, for example, 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 sound. Initially, assuming the spring is compressed to point where it's 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. #U_(sp)=U_g# Spring potential energy, #U_(sp)# is given by #1/2k(Δs)^2#. #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(9N/m)(3/5m)^2)/(0.150kg*9.8m/s^2)# #h~~1.1m#

Xavier Adame · Answer

undefined To find the maximum height reached by the ball, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the system (when the spring is compressed) is equal to the final mechanical energy of the system (when the ball reaches its maximum height).

The initial mechanical energy (E_initial) is the potential energy stored in the compressed spring, and the final mechanical energy (E_final) is the potential energy of the ball when it reaches its maximum height.

The potential energy stored in the compressed spring is given by the formula: 
$PE_{initial} = \frac{1}{2}kx^2$
where:
k = spring constant = 9 (kg)/s^2
x = compression of the spring = 3/5 m

The potential energy of the ball when it reaches its maximum height is given by the formula: 
$PE_{final} = mgh$
where:
m = mass of the ball = 0.15 kg (converted from 150 g)
g = acceleration due to gravity = 9.8 m/s^2
h = maximum height reached by the ball (to be determined)

Setting the initial mechanical energy equal to the final mechanical energy, we have:
$\frac{1}{2}kx^2 = mgh$

Now, we can solve for h:
$h = \frac{\frac{1}{2}kx^2}{mg}$

Substituting the given values, we get:
$h = \frac{\frac{1}{2} \times 9 \times (3/5)^2}{0.15 \times 9.8}$

Solving this equation gives:
$h \approx 0.275\,m$

Therefore, the maximum height reached by the ball is approximately 0.275 meters.