Question

question 8 (1 point) a 1000 - kg car is moving at 15 km/h. if a 2000 - kg truck has 23 times the kine energy of the car, how fast is the truck moving? a) 48 km/h b) 41 km/h c) 82 km/h d) 51 km/h e) 61 km/h f) 72 km/h

Explanation:

Step1: Write the kinetic energy formula

The kinetic energy formula is \(K = \frac{1}{2}mv^{2}\). Let \(m_{1}=1000\space kg\), \(v_{1} = 15\space km/h\) for the car, and \(m_{2}=2000\space kg\), \(v_{2}\) for the truck. Given \(K_{2}=23K_{1}\).

Step2: Substitute into the kinetic - energy relationship

Substitute \(K=\frac{1}{2}mv^{2}\) into \(K_{2}=23K_{1}\). We get \(\frac{1}{2}m_{2}v_{2}^{2}=23\times\frac{1}{2}m_{1}v_{1}^{2}\).
Cancel out \(\frac{1}{2}\) on both sides: \(m_{2}v_{2}^{2}=23m_{1}v_{1}^{2}\).

Step3: Solve for \(v_{2}\)

Substitute \(m_{1} = 1000\), \(m_{2}=2000\), \(v_{1}=15\) into \(m_{2}v_{2}^{2}=23m_{1}v_{1}^{2}\).
\(2000v_{2}^{2}=23\times1000\times15^{2}\).
First, simplify the equation: \(2v_{2}^{2}=23\times225\) (divide both sides by \(1000\)).
\(2v_{2}^{2}=5175\).
\(v_{2}^{2}=\frac{5175}{2}=2587.5\).
\(v_{2}=\sqrt{2587.5}\approx50.87\approx51\space km/h\).

Answer:

d) \(51\space km/h\)