A 120-kg linebacker accelerates from 5.0 m/s to 10.0 m/s in 1.0 s. how much power does that require?

Power is the rate at which work is done in a given period. Mathematically, power is the work/time ratio.

P = W/t

Description: P = power (Joule/second = Watt), W = work (Joule), t = time interval (second)

Based on this equation, it can be concluded that the greater the work rate, the greater the power. On the other hand, the smaller the work rate, the smaller the power. Work rate refers to the rate at which work is done.

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Power is a scalar. The SI unit of power is the Joule/second. The Joule/second = Watt (abbreviated as W), named so to pay homage to James Watt. The British imperial unit for power is foot-pound per second.

This unit is too small for practical purposes, so the greater unit horsepower (abbreviated as hp) is used. One horsepower = 550 foot-pound per second = 764 Watt = ¾ kilowatt.

The amount of work can also be expressed in power x time units, for instance, kilowatt-hour or kWh. One kWh refers to the work done at a constant rate of 1 kiloWatt for one hour.

1. A 50-kg person runs up the stairs 10 meters high in 2 minutes. Acceleration due to gravity (g) is 10 m/s2. Determine the power.

Known :

Mass (m) = 50 kg

Height (h) = 10 meters

Acceleration due to gravity (g) = 10 m/s2

Time interval (t) = 2 minute = 2 (60) = 120 seconds

Wanted : Power (P)

Solution :

Formula of power :

P = W / t

P = power, W = work, t = time

Formula of Work :

W = F s = w h = m g h

W = work, F = force, w = weight, d = displacement, h = height, m = mass, g = acceleration due to gravity

W = m g h = (50)(10)(10) = 5000 Joule.

P = W / t = 5000 / 120 = 41.7 Joule/second.

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2. Calculate the power required of a 60-kg person who climbs a tree 5 meters high in 10 seconds. Acceleration due to gravity is 10 m/s2.

Known :

Mass (m) = 60 kg

Height (h) = 5 meters

Acceleration due to gravity (g) = 10 m/s2

Time interval (t) = 10 seconds

Wanted : Power

Solution :

Work :

W = m g h = (60)(10)(5) = 3000 Joule

Power :

P = W / t = 3000 / 10 = 300 Joule/second.

3. A rotary comedy with a power of 300 watts and period of 5 minutes rotates 5 rounds. The energy it uses is ….

A. 15 kJ

B. 75 kJ

C. 90 kJ

D. 450 kJ

Known :

Power (P) = 300 Watt = 300 Joule/second

Period (T) = 5 minutes = 5 (60 seconds) = 300 seconds

Number of rotation = 5

Wanted: Energy used by the rotary comedy

Solution :

The correct answer is D.

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