Force, Work, and Power - a Review

By Jeff Lucius

If an object is being accelerated, then force is being applied in the same direction as the acceleration. If force is being applied to an object that does not move then an equal but opposite force must also be applied. Force, F, is then defined as

F = m x a,

where m = mass, kilogram (kg) in SI (international system of units) or slug in the engineering system, and a = acceleration, m/s2 in SI or ft/s2 in the engineering system. One slug equals 32.17 pounds mass or 14.59 kilograms. One foot exactly equals 0.3048 meter. The newton, N, is the unit of force in SI (kg-m/s2). The pound force (lbf) is the unit of force in the engineering system. One pound force equals 4.4482216 newtons. Force is a vector quantity; the direction of the applied force is as important as the magnitude.

If a force, F, tends to rotate a body about an axis, then that tendency is measured by the magnitude of the force times the perpendicular distance, l, between the line of action of the force and the axis. The distance, l, is called the moment arm of the force and the product of F and l is called the moment or torque, G, of the force. The units of torque are either newton-meter (N-m) or pound force-foot (lbf-ft).

G = F x l

Work is being done if there is motion of an object along some component direction of the force (not at right angles). Work, W, can then be defined as

W = F x cos q x d,

where d = the linear displacement of the object, and q (Theta) = the angle between the direction of displacement and the direction the force is applied. The magnitude of the force in the direction of displacement is F x cosq. When displacement is measured in meters then the unit of work is the joule (J) which equals one newton-meter (N-m). If displacement is measured in feet, then the unit of work is the foot-pound force (ft-lbf), which should not be confused with the unit of torque, the lbf-ft. One joule equals 0.7376 ft-lbf. Work is a scalar quantity, which means it is defined only by its magnitude. If a force is acting to rotate a body, then work is redefined in terms of torque and angular displacement, q.

W = G x q

Power is the rate at which work is performed. When an object is moved linearly, power, P, is defined as the product of force, applied in the direction of movement, and the linear velocity;

P = W / time = F x v,

where v = velocity. In SI, power is measured in watts, W. One watt equals the application of one joule per second; W = J/s = N-m/s. In the engineering system, the unit of power is the horsepower, HP,

1 HP = 550 ft-lbf/s = 33,000 ft-lbf/min = 746 W.

Power also can be defined as the product of torque and the instantaneous angular velocity, w.

P = W / time = G x w,

where P = power measured in lbf-ft/min, G (Gamma) = torque measured in lbf-ft, and w (omega) = instantaneous angular velocity in radians per minute. Instantaneous angular velocity is the product of the RPM (revolutions per minute) and 2p (2 pi) radians (3600) per revolution.

Analogy between translational and rotational motion
Concept Translation Rotation Comments
Displacement x q x = rq
Velocity v = dx/dt w  = dq/dt v = rw
Acceleration a = dv/dt α = dw/dt  
Force, moment F G G  = rF
Equilibrium ΣF = 0 ΣG  = 0  
Constant acceleration v = v0 + at w  = w0 + αt  
x = x0 + v0t + ½at2 q = q0 + w0t + ½αt2  
v2 = v02 + 2a(x-x0) w2 = w02 + 2α(q-q0)  
Mass, moment of inertia m I I = Σmiri2
Newton's second law ΣF = ma ΣG = Iα  
Work W = ∫F dx W = ∫G dq  
Power P = Fv P = Gw  
Kinetic energy K = ½mv2 K = ½Iw2  
Impulse ∫F dt G dt  
Momentum mv L = Iw  

To quantify the performance of an automobile, we often cite the torque and HP of the engine and values for acceleration or top speed. Engine torque is the amount of force applied to rotate the crankshaft or flywheel. Technically, lots of torque can be applied without rotating an object such as the flywheel. Fortunately for us, engine torque is accompanied by lots of "wheel spin". Engine horsepower (or power) is the torque at the flywheel times the angular velocity of the flywheel. There is a simple expression relating HP and torque.

HP x 33,000 ft-lbf/min = torque (lbf-ft) x RPM (revolutions/min) x 2p (radians/revolution)
HP = (torque x RPM x 6.283185) ¸ 33,000
HP = torque x RPM ¸ 5252.113.

It is clear from this relationship that the magnitude of HP will always be less than the magnitude of torque below 5252 RPM, that the magnitudes will be equal at ~5252 RPM, and that HP magnitude will be greater than torque magnitude above ~5252 RPM, remembering their respective units, of course.

Maximum horsepower can be estimated crudely from a carís mass and quarter mile elapsed time and terminal velocity.

Power = W / time = F x v = F x d ¸ t = m x a x d ¸ t.

I will use the weight of my car as an example and Jack Tertadian's (nitrous-assisted) June 1997 ľ mile ET of 10.81 seconds and terminal velocity of 128.44 mph. Average acceleration is the difference in velocity divided by the difference in time. Here this would be 128.44 mph ¸ 10.81 s. Converting 128.44 mph to ft/s gives 188.4 ft/s (128.44 x 5280 ¸ 3600) and an average acceleration of 17.43 ft/s2 (188.4 ¸ 10.81) or 5.31 m/s2. My car weighs about 4000 lb (1814 kg) with driver, fuel, etc.; the ľ mile is 402.336 m long. It is easier to calculate power in the SI system first then convert to HP.

Power = 1814 kg x 5.31 m/s2 x 402.336 m ¸ 10.81 s = 358505 (kg-m/s2) x m/s = 358505 N-m/s = 358505 W.

Dividing by 746 watts per HP we get about 480 HP at the wheels. Assuming about 110 HP loss from flywheel to drive wheels (109 HP @ 6000 RPM with 362 HP output in 4th gear, according to Jim Mathews) the engine would produce about 590 HP. This figure does not include aerodynamic drag "losses". These might be about 50 HP at the wheels for 100 mph and maybe 100 HP at the wheels for 125 MPH. Adding in aerodynamic drag brings engine power to near 700 HP. Incredible. Accelerometers, such as G Techä Performance Meter/Pro, calculate horsepower in a similar manner. A chassis dynamometer is required to calculate the actual HP at the wheels as a function of RPM.

As another example, the constant acceleration formula can be used to determine the approximate distance a car has traveled after accelerating from a dead stop to 60 mph. In the formula, x = x0 + v0t + ½at2, x0 and v0 are zero. So, the distance traveled after reaching 60 mph is calculated using the expression ½at2. Let's assume the car reaches 6o mph in 4 seconds. The acceleration, a, is 22 ft/s2, which is 60 mph or 88 ft/s (60 mph times 5280 feet per mile divided by 3600 s) divided by 4 seconds. Divide that by 2 to get 11 ft/s2. Finally multiply 11 ft/s2 times 4 seconds squared (16 s2) to get 176 feet traveled.

So, how fast would a car have to accelerate to 60 mph to achieve 1 g (32 ft/s2) of acceleration? The time would be 88 ft/s divided by 32 ft/s2 or 2.75 seconds.

The formulas in the table above are useful to quantify performance as well as determine, for example, the kinetic energy stored in a flywheel, the moment of inertia (rotating mass) of drivetrain parts, or the work required by the brakes to stop a vehicle.

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Page last updated February 15, 2008