A Spring K80 N·m Used To Propel A 150 Gram Steel Ball
A Spring K80 Nm Is Used To Propel A 150 Gram Steel Ball Along A Tr
A spring (k=80 N/m) is used to propel a 150 gram steel ball along a track that includes a ramp inclined 32 degrees above the vertical. The ball is pushed to the left, compressing the spring, then released to roll along the track. Friction is neglected. The problem asks for the spring compression distance before release, the maximum vertical height reached on the ramp, and the linear distance traveled along the ramp.
Paper For Above instruction
This problem involves classical mechanics principles, specifically energy conservation, kinematics, and dynamics of rolling objects. The main tasks include calculating the initial compression of the spring based on the kinetic energy of the ball after release, determining the maximum height the ball reaches along the inclined track, and finding the total distance traveled along the ramp before rolling back.
Part a: Determining the spring compression distance
Initially, when the spring is compressed, it stores elastic potential energy, which is converted into the kinetic energy of the rolling ball upon release. Given the spring constant \(k=80 \, \mathrm{N/m}\) and the mass of the steel ball \(m=150\, \mathrm{g} = 0.150\, \mathrm{kg}\), with the final velocity \(v=3.75\, \mathrm{m/s}\). Assuming no energy losses, the elastic potential energy stored in the spring is equal to the kinetic energy of the ball after release:
\[
PE_{spring} = KE_{ball}
\]
The elastic potential energy of the spring when compressed by distance \(x\):
\[
PE_{spring} = \frac{1}{2} k x^2
\]
The kinetic energy of the rolling ball:
\[
KE_{ball} = \frac{1}{2} m v^2
\]
Setting these equal:
\[
\frac{1}{2} k x^2 = \frac{1}{2} m v^2
\]
Simplifying to solve for \(x\):
\[
x = \sqrt{\frac{m v^2}{k}}
\]
Plugging in the values:
\[
x = \sqrt{\frac{(0.150\, \mathrm{kg})(3.75\, \mathrm{m/s})^2}{80\, \mathrm{N/m}}}
\]
\[
x = \sqrt{\frac{(0.150)(14.0625)}{80}} = \sqrt{\frac{2.109375}{80}} = \sqrt{0.0263671875}
\]
\[
x \approx 0.162\, \mathrm{m}
\]
Thus, the spring was compressed approximately 0.162 meters (16.2 cm) before release.
Part b: Maximum vertical height reached on the ramp
As the ball rolls up the inclined track, its kinetic energy converts into potential energy until it reaches maximum height where kinetic energy is zero. The initial kinetic energy immediately after release is:
\[
KE = \frac{1}{2} m v^2
\]
At the maximum height, all this kinetic energy is converted into gravitational potential energy:
\[
PE_{gravity} = m g h_{max}
\]
Equate the two:
\[
\frac{1}{2} m v^2 = m g h_{max}
\]
\[
h_{max} = \frac{v^2}{2 g}
\]
Using \(v=3.75\, \mathrm{m/s}\) and \(g=9.81\, \mathrm{m/s^2}\):
\[
h_{max} = \frac{(3.75)^2}{2 \times 9.81} = \frac{14.0625}{19.62} \approx 0.717\, \mathrm{m}
\]
This height is measured relative to the point of release. However, since the track is inclined at 32°, the vertical height can also be expressed as a component along the incline but the maximum height above the initial position is approximately 0.717 meters.
Part c: Linear distance along the ramp
The total distance along the ramp that the ball travels before coming to rest (assuming it turns around at maximum height) corresponds to the length of the inclined path where the vertical height is maximum. This length \(L\) can be found using the relation:
\[
h_{max} = L \sin \theta
\]
where \(\theta = 32^\circ\).
Rearranged:
\[
L = \frac{h_{max}}{\sin \theta}
\]
Substituting the known values:
\[
L = \frac{0.717}{\sin 32^\circ} = \frac{0.717}{0.5299} \approx 1.353\, \mathrm{m}
\]
Therefore, the ball travels approximately 1.35 meters along the inclined ramp before reaching its maximum height and rolling back down.
Summary of Results:
- Spring compression distance: 0.162 meters
- Maximum vertical height: 0.717 meters
- Distance traveled along the ramp: 1.35 meters
These calculations are based on energy conservation principles, assuming no friction losses, and ideal rolling without slipping.
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