Question

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a vertical spring is mounted to a table and compressed downward. a ball of mass ( m ) is placed on a spring (spring constant ( k ) ), the spring is released, and the ball is projected vertically upward. in which case would you expect the ball to reach the greatest maximum height?
compress the spring twice as much
increase the spring constant by a factor of two
reduce the mass of the ball by a factor of two
decrease the spring constant by a factor of two

Explanation:

Step1: Apply conservation of energy

The elastic potential energy of a spring is \(U = \frac{1}{2}kx^{2}\), and at the maximum height, all the elastic potential energy is converted into gravitational potential energy \(U_{g}=mgh\). So, \(\frac{1}{2}kx^{2}=mgh\), and \(h=\frac{kx^{2}}{2mg}\).

Step2: Analyze each option

• **Compress the spring twice as much (\(x

ightarrow2x\))**:
Substitute \(x = 2x\) into \(h=\frac{kx^{2}}{2mg}\), we get \(h_{1}=\frac{k(2x)^{2}}{2mg}=\frac{4kx^{2}}{2mg} = 4h\).

• **Increase the spring constant by a factor of two (\(k

ightarrow2k\))**:
Substitute \(k = 2k\) into \(h=\frac{kx^{2}}{2mg}\), we get \(h_{2}=\frac{(2k)x^{2}}{2mg}=2h\).

• **Reduce the mass of the ball by a factor of two (\(m

ightarrow\frac{m}{2}\))**:
Substitute \(m=\frac{m}{2}\) into \(h=\frac{kx^{2}}{2mg}\), we get \(h_{3}=\frac{kx^{2}}{2(\frac{m}{2})g}=2h\).

• **Decrease the spring constant by a factor of two (\(k

ightarrow\frac{k}{2}\))**:
Substitute \(k=\frac{k}{2}\) into \(h=\frac{kx^{2}}{2mg}\), we get \(h_{4}=\frac{(\frac{k}{2})x^{2}}{2mg}=\frac{kx^{2}}{4mg}=\frac{h}{2}\).

Answer:

Compress the spring twice as much.