fbpx

Power – problems and solutions

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.

See also  Electric charges Magnetic field and Magnetic forces – Problems and Solutions

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.

See also  Vector displacement – problems and solutions

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 :

Power – problems and solutions

The correct answer is D.

[wpdm_package id=’1190′]

  1. Work was done by force problems and solutions
  2. Work-kinetic energy problems and solutions
  3. Work-mechanical energy principle problems and solutions
  4. Gravitational potential energy problems and solutions
  5. The potential energy of elastic spring problems and solutions
  6. Power problems and solutions
  7. Application of conservation of mechanical energy for free fall motion
  8. Application of conservation of mechanical energy for up and down motion in free fall motion
  9. Application of conservation of mechanical energy for motion on a curved surface
  10. Application of conservation of mechanical energy for motion on an inclined plane
  11. Application of conservation of mechanical energy for projectile motion

Print Friendly, PDF & Email

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Discover more from Physics

Subscribe now to keep reading and get access to the full archive.

Continue reading