Question

a spring with k = 150n/m is stretched 0.3 m. how much energy does it store?
kinetic energy (ke)
a 2 kg ball rolls at 3 m/s. find its kinetic energy.
a 1.5 kg toy car moves at 4 m/s. calculate its kinetic energy.
a 0.5 kg baseball is thrown at 20 m/s. how much kinetic energy does it have?
a 1000 kg car travels at 15 m/s. find its kinetic energy.

Explanation:

Step1: Elastic potential energy formula

The formula for elastic potential energy \(E_{p}=\frac{1}{2}kx^{2}\), where \(k = 150N/m\) and \(x=0.3m\).
Substitute the values: \(E_{p}=\frac{1}{2}\times150\times(0.3)^{2}\)

Step2: Calculate the value

First, calculate \((0.3)^{2}=0.09\). Then \(\frac{1}{2}\times150\times0.09 = 75\times0.09=6.75J\)

Step3: Kinetic energy formula

The formula for kinetic energy \(KE=\frac{1}{2}mv^{2}\)

For the \(2kg\) ball:

\(m = 2kg\), \(v = 3m/s\). Substitute into the formula: \(KE=\frac{1}{2}\times2\times(3)^{2}\)
Calculate \((3)^{2}=9\), then \(\frac{1}{2}\times2\times9=9J\)

For the \(1.5kg\) toy - car:

\(m = 1.5kg\), \(v = 4m/s\). Substitute: \(KE=\frac{1}{2}\times1.5\times(4)^{2}\)
\((4)^{2}=16\), \(\frac{1}{2}\times1.5\times16 = 1.5\times8 = 12J\)

For the \(0.5kg\) baseball:

\(m=0.5kg\), \(v = 20m/s\). Substitute: \(KE=\frac{1}{2}\times0.5\times(20)^{2}\)
\((20)^{2}=400\), \(\frac{1}{2}\times0.5\times400=0.25\times400 = 100J\)

For the \(1000kg\) car:

\(m = 1000kg\), \(v = 15m/s\). Substitute: \(KE=\frac{1}{2}\times1000\times(15)^{2}\)
\((15)^{2}=225\), \(\frac{1}{2}\times1000\times225=500\times225 = 112500J\)

Answer:

• The spring stores \(6.75J\) of energy.
• The \(2kg\) ball has \(9J\) of kinetic energy.
• The \(1.5kg\) toy - car has \(12J\) of kinetic energy.
• The \(0.5kg\) baseball has \(100J\) of kinetic energy.
• The \(1000kg\) car has \(112500J\) of kinetic energy.