Interactive physics simulator
Power: P = W/t
Explore physical power as the rate of performing mechanical work. Investigate weightlifting kinetics, elevator electrical motor efficiency, and electric car drag strip runs in three high-fidelity simulation environments.
Power & Work Rate Lab
Adjust masses, vertical dimensions, elapsed times, and efficiencies to visualize mechanical work rate conversions.
Live Telemetry
- Barbell Mass (m)
- 150 kg
- Target Height (h)
- 2.0 m
- Lift Duration (t)
- 1.5 s
- Work Done (W)
- 0 J
- Lifting Force (F)
- 0 N
- Average Power (P)
- 0 W
- Horsepower (hp)
- 0.00 hp
- Passenger Load (m)
- 800 kg
- Shaft Height (h)
- 30 m
- Ascent Time (t)
- 12.0 s
- Efficiency (η)
- 80%
- Useful Output Power
- 0 kW
- Electrical Input Power
- 0 kW
- Wasted Heat Loss
- 0 kJ
- Car Mass (m)
- 1600 kg
- Target Speed (v)
- 30.0 m/s
- Acceleration Time (t)
- 4.0 s
- Kinetic Energy (W)
- 0 kJ
- Tractive Force (F)
- 0 N
- Average Power
- 0 kW
- Peak Instantaneous Power
- 0 kW
What is Power?
In physics, power is defined as the rate at which work is done or energy is transferred. While work measures the total change in energy, power measures how quickly that change occurs. Mathematically, average power is expressed as:
Where:
- P is the average power output.
- W is the total work done (or energy transferred) in Joules (J).
- t is the time duration in seconds (s) over which the work is performed.
SI Units & Horsepower
The standard SI unit of power is the Watt (W), named in honor of James Watt. One Watt represents a rate of energy conversion of one Joule per second:
In engineering, particularly when discussing motors, vehicles, and combustion engines, the unit Horsepower (hp) is frequently employed. The mechanical horsepower is defined as:
Power, Force, and Velocity
For an object moving in a straight line under a constant force, we can relate power directly to speed. Since work is force times displacement (W = F * d), we can substitute this into the power equation:
This reveals that instantaneous power (P) is the product of the force (F) acting on an object and its instantaneous velocity (v) in the direction of the force:
This relationship is crucial for understanding vehicle mechanics. It demonstrates why high-speed operation against frictional resistances requires massive power outputs.
Power and Efficiency
No real-world motor converts electrical or chemical energy to mechanical work with 100% efficiency. Friction, heat dissipation, and electromagnetic resistance convert a portion of the input power into thermal waste. The efficiency eta of a system is defined as:
Where:
- Pout is the useful power output delivered by the system to do work (e.g., pulling a load).
- Pin is the total power input drawn from the energy source (e.g., electrical grid).
- The rate of energy waste (thermal dissipation) is given by: Pwaste = Pin - Pout.
Solved Numerical Examples
A weightlifter lifts a mass m = 150 kg from the floor to a vertical height of h = 2.0 meters in a duration of t = 1.5 seconds. Calculate: (a) the work done against gravity, and (b) the average power output of the lifter in Watts and Horsepower. Use g = 9.8 m/s².
View Step-by-Step Solution
- Identify the given values: mass m = 150 kg, height h = 2.0 m, lift duration t = 1.5 s, and gravity g = 9.8 m/s².
- Calculate the force required to lift the barbell at a constant speed, which is equal to the gravitational weight: F = m · g = 150 · 9.8 = 1,470 Newtons.
- Calculate the mechanical work done: W = F · h = m · g · h = 1,470 · 2.0 = 2,940 Joules.
- Apply the power formula P = W / t to find the average power output: P = 2,940 / 1.5 = 1,960 Watts.
- Convert the power output to Horsepower (hp) using the conversion factor 1 hp ≈ 746 Watts: P_hp = 1,960 / 746 ≈ 2.63 hp.
An electric elevator hoist with an efficiency of 80% lifts a passenger cabin of mass m = 800 kg to a vertical height of h = 30 meters in t = 12.0 seconds at a constant speed. Calculate: (a) the useful work output, (b) the useful power output, (c) the total electrical power input drawn by the motor, and (d) the electrical energy wasted as thermal dissipation. Use g = 9.8 m/s².
View Step-by-Step Solution
- Identify the given values: cabin mass m = 800 kg, vertical height h = 30 m, ascent duration t = 12.0 s, efficiency η = 0.80 (80%), and gravity g = 9.8 m/s².
- Calculate the useful work output done in lifting the elevator cabin: W_out = m · g · h = 800 · 9.8 · 30 = 235,200 Joules = 235.2 kJ.
- Calculate the useful mechanical power output delivered to the cable hoist: P_out = W_out / t = 235,200 / 12.0 = 19,600 Watts = 19.6 kW.
- Use the efficiency equation (η = P_out / P_in) to solve for the electrical power input: P_in = P_out / η = 19,600 / 0.80 = 24,500 Watts = 24.5 kW.
- Calculate the total electrical energy input supplied to the motor: W_in = P_in · t = 24,500 · 12.0 = 294,000 Joules = 294.0 kJ.
- Determine the electrical energy wasted as heat due to electromagnetic and frictional friction: E_heat = W_in - W_out = 294,000 - 235,200 = 58,800 Joules = 58.8 kJ.
A high-performance electric vehicle (EV) of mass m = 1600 kg accelerates from rest to a speed of v = 30 m/s (approx. 108 km/h) in a time interval of t = 4.0 seconds on a flat drag strip. Assuming constant acceleration and neglecting air resistance, calculate: (a) the mechanical work done to accelerate the car, (b) the average power output, and (c) the peak instantaneous power delivered by the motor at the end of the 4.0 seconds.
View Step-by-Step Solution
- Identify the given values: vehicle mass m = 1,600 kg, target speed v_final = 30 m/s, and acceleration duration t = 4.0 s.
- According to the Work-Energy Theorem, the net work done equals the change in kinetic energy: W = ΔKE = 1/2 · m · v_final² - 0 = 0.5 · 1,600 · 30² = 720,000 Joules = 720 kJ.
- Calculate the average power output over the 4.0-second acceleration period: P_avg = W / t = 720,000 / 4.0 = 180,000 Watts = 180 kW (approx. 241.3 hp).
- For a constant accelerating force, acceleration is constant: a = v_final / t = 30 / 4.0 = 7.5 m/s². The net force is F = m · a = 1,600 · 7.5 = 12,000 Newtons.
- Calculate the peak instantaneous power output at the exact moment the vehicle reaches its final speed (v_final = 30 m/s): P_peak = F · v_final = 12,000 · 30 = 360,000 Watts = 360 kW (approx. 482.6 hp).
- Note that for constant acceleration from rest, the instantaneous speed increases linearly, meaning the instantaneous power P(t) = F · a · t also increases linearly. The peak power at the end of the run is exactly twice the average power: P_peak = 2 · P_avg = 2 · 180 kW = 360 kW.
Conceptual Practice
Why does running up a flight of stairs require a much higher power output than walking up the same stairs, even though the total work done is identical?
Show Explanation
Work is the product of force and distance. In both cases, the vertical distance (height h) and the force exerted (weight equal to mg) are the same, so the total mechanical work done against gravity is W = mgh. However, power is the rate of doing work (P = W/t). Running reduces the elapsed time (t) to perform this work. Because the same amount of work is accomplished in a shorter time interval, the power output when running is significantly greater than when walking.
Explain why doubling the power output of a sports car engine does not double its maximum top speed.
Show Explanation
At top speed, a vehicle is no longer accelerating, meaning the engine force matches resistive forces. The primary resistance at high speeds is aerodynamic drag, which is proportional to the square of velocity (F_drag ∝ v²). Since power is the product of force and velocity (P = F · v), the power required to overcome drag scales with the cube of velocity (P_required ∝ v³). Consequently, doubling the power increases the maximum top speed by a factor of ∛2 ≈ 1.26, which is only a 26% speed increase.
In a continuous mechanical system like an elevator hoist, how does lowering the ascent speed affect the motor power and cable tension?
Show Explanation
Assuming constant ascent velocity, the cable tension is equal to the downward weight of the cabin (T = mg) to maintain equilibrium, meaning the force is independent of speed. However, because power is force times velocity (P = T · v), the required power is directly proportional to speed. Halving the speed cuts the required motor power in half, allowing a less powerful motor to lift the load, although it takes twice as long.
Why is the peak instantaneous power exactly twice the average power when a vehicle accelerates from rest under constant force?
Show Explanation
Under constant force F, the vehicle accelerates at a constant rate a, causing velocity to increase linearly over time: v(t) = a · t. Since instantaneous power is P(t) = F · v(t), power also increases linearly over time from 0 to a maximum value of P_peak = F · v_final. The average value of a linear function starting at zero is exactly half of its peak value: P_avg = (0 + P_peak) / 2 = 1/2 · P_peak. Therefore, the peak power at the end of the run is exactly twice the average power.
Frequently Asked Questions
What is power in physics?
Power (P) is the rate at which work is done or energy is transferred. It measures how fast energy is used or work is accomplished, defined mathematically as P = W/t where W is work and t is time.
What is the SI unit of power?
The SI unit of power is the Watt (W), named after Scottish engineer James Watt. One Watt is equal to one Joule of energy transferred or work done per second (1 W = 1 J/s = 1 kg·m2/s3).
What is Horsepower and how is it related to Watts?
Horsepower (hp) is a non-SI unit of power representing the rate at which an average draft horse could do mechanical work. One mechanical horsepower is defined as approximately 746 Watts (1 hp ≈ 746 W).
How do work and power differ?
Work is the total amount of energy transferred by a force acting over a distance, measured in Joules (J). Power is the speed or rate at which that work is done, measured in Watts (W). Doing the same work in half the time requires twice the power.
How is power related to force and velocity?
For continuous linear motion, power is the product of the force (F) acting on an object and its velocity (v) in the direction of the force: P = F · v. This is derived from P = W/t = (F · d)/t = F · (d/t) = F · v.
Why do two people of different masses climbing the same stairs at the same speed exert different power?
The heavier person must do more work against gravity because their weight (Fg = mg) is larger. Since they climb at the same speed (taking the same time t), the heavier person does more work in the same time, thereby exerting greater power.
How is the efficiency of a power system calculated?
Efficiency (η) is the ratio of useful power output to total power input: η = (Pout / Pin) × 100%. Because of friction and heat dissipation, efficiency is always less than 100% in real systems.
What does kilowatt-hour (kWh) measure, power or energy?
A kilowatt-hour measures energy, not power. It represents the energy consumed by a 1-kilowatt device running continuously for one hour (1 kWh = 1000 W × 3600 s = 3.6 × 106 Joules).
Why does accelerating a car faster require a more powerful engine?
Accelerating a car increases its kinetic energy. Reaching a target kinetic energy (KE = 1/2mv2) in a shorter time (t) requires a higher rate of work, which means the engine must have a larger power output (P = change in KE / t).
What is the difference between average power and instantaneous power?
Average power is the total work done divided by the total time interval (Pavg = W / t). Instantaneous power is the rate of doing work at a specific single point in time, calculated as the limit of average power as the time interval approaches zero, or P = F * v at that instant.
How does elevator load affect motor power requirements?
A heavier elevator cabin requires a larger tension force in the cables to lift it against gravity. For a constant ascent speed v, the motor must supply a larger force (F = mg), requiring more power output (P = F * v).
Why do heavy trucks have high-power engines but low top speeds compared to sports cars?
Truck engines are geared to produce massive forces (torque) to move heavy loads, meaning F is very large but v is low. Sports cars are optimized for high speed (v is large) with lower mass and force requirements. Both can have high power ratings, but serve different torque-to-speed ratios since P = F * v.