Question

The thrust of a certain boat’s engine generates a power of 10kW as the boat moves at constant speed 10ms through the water of a lake. The magnitude of the drag force that is exerted on the boat’s hull as it is moving through the water is directly proportional to the boat's speed and is given by the equation F=kv. The increase in power needed for the boat to move through the lake at a constant speed of 12ms is : a. 0W b. 1440 W c. 4400 W 10,000 W e. 4,400 W

Answer 1

Answer:

The change in power is 4400 W.

Explanation:

Given that,

Power = 10 kW

Speed = 10 m/s

Increases speed = 12 m/s

Given equation is,

$F=kv$

We know that,

The power is,

$P=Fv$

Put the value of F into the formula

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

We need to calculate the new power

Using formula for power

$\dfrac{P}{P'}=\dfrac{v^2}{v'^2}$

Put the value into the formula

$\dfrac{10}{P'}=(\dfrac{10}{12})^2$

$P'=(\dfrac{12}{10})^2\times10$

$P'=14.4\ kW$

We need to calculate the change in power

Using formula of change in power

$\Delta P=P'-P$

Put the value into the formula

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Hence, The change in power is 4400 W.