Question

1. what is power? what is the unit of power? define this unit in terms of kg, m, and s.

1. how much power is exerted if you lift a 200 kg barbell to a height of 2.0 m in 1.5 s?

1. two football players are running up the bleachers at the football field. larry has a mass of 120 kg and hector has a mass of 85 kg. the bleachers have a height of 35 m.

a) calculate the work done by each football player. who does more work in running up the bleachers?

b) if larry makes it in 8.50 s and hector takes 12.75 s, calculate the power exerted by each player.

1. how much work can an engine do in 60.0 s if it is capable of generating 420 horsepower? (1 hp = 746 w)

Explanation:

Question 8

Step1: Define Power

Power is the rate at which work is done or energy is transferred. Mathematically, it is defined as \( P=\frac{W}{t} \), where \( W \) is work and \( t \) is time.

Step2: Unit of Power

The SI unit of power is the watt (W).

Step3: Define Watt in SI Base Units

Work \( W \) has units of force \( F \) times distance \( d \), i.e., \( W = F\times d \). Force \( F \) has units of mass \( m \) times acceleration \( a \), and acceleration \( a \) has units of \( \frac{\text{length}}{\text{time}^2} \) (or \( \frac{m}{s^2} \) in SI). So, work \( W \) has units \( kg\times\frac{m}{s^2}\times m=kg\cdot m^2/s^2 \). Power \( P=\frac{W}{t} \), so the unit of power (watt) is \( \frac{kg\cdot m^2/s^2}{s}=kg\cdot m^2/s^3 \).

Step1: Recall the Formula for Power

Power \( P \) is given by the work done \( W \) divided by time \( t \), \( P=\frac{W}{t} \). The work done in lifting an object is equal to the gravitational potential energy gained, \( W = mgh \), where \( m \) is mass, \( g = 9.8\ m/s^2 \) (acceleration due to gravity), and \( h \) is height.

Step2: Calculate Work Done

First, calculate the work done in lifting the barbell. \( m = 200\ kg \), \( g = 9.8\ m/s^2 \), \( h = 2.0\ m \). So, \( W=mgh=200\ kg\times9.8\ m/s^2\times2.0\ m \).
\( W = 200\times9.8\times2= 3920\ J \) (since \( 1\ J = 1\ kg\cdot m^2/s^2 \)).

Step3: Calculate Power

Now, use \( P=\frac{W}{t} \), with \( t = 1.5\ s \). So, \( P=\frac{3920\ J}{1.5\ s}\approx2613.33\ W \) (or we can also write it as approximately \( 2.6\times 10^3\ W \) or \( 2.6\ kW \)).

Step1: Recall the Formula for Work Done Against Gravity

The work done \( W \) in lifting (or moving up against gravity) an object is given by \( W = mgh \), where \( m \) is mass, \( g = 9.8\ m/s^2 \), and \( h \) is the vertical height.

Step2: Calculate Work Done by Larry

For Larry, \( m_{Larry}=120\ kg \), \( h = 35\ m \), \( g = 9.8\ m/s^2 \). So, \( W_{Larry}=m_{Larry}gh=120\ kg\times9.8\ m/s^2\times35\ m \).
\( W_{Larry}=120\times9.8\times35 = 120\times343 = 41160\ J \).

Step3: Calculate Work Done by Hector

For Hector, \( m_{Hector}=85\ kg \), \( h = 35\ m \), \( g = 9.8\ m/s^2 \). So, \( W_{Hector}=m_{Hector}gh=85\ kg\times9.8\ m/s^2\times35\ m \).
\( W_{Hector}=85\times9.8\times35 = 85\times343 = 29155\ J \).

Step4: Compare the Work Done

Since \( 41160\ J>29155\ J \), Larry does more work.

Answer:

• Power: The rate of doing work (or transferring energy), \( P = \frac{W}{t} \).
• Unit of power: Watt (W).
• Definition in terms of \( kg, m, s \): \( 1\ W=1\ kg\cdot m^2/s^3 \) (since \( P=\frac{W}{t} \), \( W = Fd=mad \), \( F = ma \), \( a=\frac{m}{s^2} \), so \( W=kg\times\frac{m}{s^2}\times m = kg\cdot m^2/s^2 \), then \( P=\frac{kg\cdot m^2/s^2}{s}=kg\cdot m^2/s^3 \)).

Question 9