Question

w=δe = e_f - e_i
= ½mv_f² - ½mvi² = ½m(v_f² - v_i²)

• a catamaran with a mass of 5.44×103 kg is moving at 12 knots. how much work is required to increase the speed to 16 knots? (one knot = 0.51 m/s.)
• an 875.0-kg car speeds up from 22.0 m/s to 44.0 m/s. what are the initial and final kinetic energies of the car? how much work is done on the car to increase its speed?

Explanation:

First Problem (Catamaran)

We use the work - kinetic energy theorem \(W=\Delta E_{k}=\frac{1}{2}m(v_{f}^{2}-v_{i}^{2})\). First, we need to convert the speed from knots to m/s. Given that 1 knot = 0.51 m/s.

Step 1: Convert speeds to m/s

• Initial speed \(v_{i}=12\) knots. So \(v_{i}=12\times0.51\ m/s = 6.12\ m/s\)
• Final speed \(v_{f}=16\) knots. So \(v_{f}=16\times0.51\ m/s=8.16\ m/s\)
• Mass of the catamaran \(m = 5.44\times10^{3}\ kg\)

Step 2: Calculate the work done

Using the formula \(W=\frac{1}{2}m(v_{f}^{2}-v_{i}^{2})\)
\[

$$\begin{align*} W&=\frac{1}{2}\times5.44\times 10^{3}\times(8.16^{2}-6.12^{2})\\ &=\frac{1}{2}\times5.44\times 10^{3}\times(66.5856 - 37.4544)\\ &=\frac{1}{2}\times5.44\times 10^{3}\times29.1312\\ &=2.72\times 10^{3}\times29.1312\\ &=79236.864\ J\approx7.92\times 10^{4}\ J \end{align*}$$

\]

Second Problem (Car)

Part 1: Initial Kinetic Energy (\(E_{ki}\))

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

Step 1: Calculate initial kinetic energy

Given \(m = 875.0\ kg\) and \(v_{i}=22.0\ m/s\)
\[

$$\begin{align*} E_{ki}&=\frac{1}{2}\times875.0\times(22.0)^{2}\\ &=\frac{1}{2}\times875.0\times484\\ & = 875.0\times242\\ &=211750\ J = 2.12\times 10^{5}\ J \end{align*}$$

\]

Part 2: Final Kinetic Energy (\(E_{kf}\))

Using the kinetic energy formula \(E_{k}=\frac{1}{2}mv^{2}\)

Step 1: Calculate final kinetic energy

Given \(m = 875.0\ kg\) and \(v_{f}=44.0\ m/s\)
\[

$$\begin{align*} E_{kf}&=\frac{1}{2}\times875.0\times(44.0)^{2}\\ &=\frac{1}{2}\times875.0\times1936\\ &=875.0\times968\\ & = 847000\ J=8.47\times 10^{5}\ J \end{align*}$$

\]

Part 3: Work done to increase speed

Answer:

s:

• Work required for catamaran: \(\boldsymbol{\approx7.92\times 10^{4}\ J}\)
• Initial kinetic energy of car: \(\boldsymbol{2.12\times 10^{5}\ J}\)
• Final kinetic energy of car: \(\boldsymbol{8.47\times 10^{5}\ J}\)
• Work done on car: \(\boldsymbol{6.35\times 10^{5}\ J}\)