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Work Against Gravity and Inertia by an External Force
by Ron Kurtus (3 December 2009)
Work against gravity is achieved by applying a sufficient external force to move an object a certain distance in the opposite direction of gravity. The work is the product of the force and the distance moved.
If the object is initially stationary, the applied upward force must overcome both inertia and gravity to move the object
a distance. If the object is moving upward at some velocity, a force only equal to gravity will continue the upward movement
at that velocity.
There are two common situations for determining the work required. You can project an object upward to a given height, where you let it continue to move—similar to throwing an object upward. In the other situation, you move the object upward to a given velocity, keep it at that velocity and then cause it to slow down to zero velocity—as done when you lift something.
Questions you may have include:
- What is the work done against gravity and inertia?
- How is inertia a factor?
- How much energy is required to do the work?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Work while accelerating object
When you apply a sufficient upward force to a stationary object and continue to accelerate it until you reach a given height, you must do work against both gravity and inertia for the whole distance. After you stop applying the force, you are no longer doing work. However, the object will then move freely upward, slowing down because of gravity.
Examples of this are propelling a rocket upward and throwing a ball upward. In both cases, the object is accelerated for a certain distance before it is allowed to move freely. Work is determined for the distance covered in the acceleration.
Force to move upward
When you push a stationary object upward with the force of F = mg, the object will not move because the upward and downward forces neutralize each other. Instead, the force required to move a stationary object upward is the sum of the force to overcome resistance from inertia and an upward force equal to that of gravity:
Fg = mg
Fi = ma
F = Fg + Fi
where
- Fg is the upward force to overcome the gravity in newtons (N) or pound-force (lbs)
- g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
- Fi is the force required to overcome the inertia in newtons (N) or pounds-force (lbs)
- m is the mass of the object in kilograms (kg) or pounds-mass
- a is the upward acceleration of the object in meters per second-squared (m/s2) or feet per second-squared (ft/s2)
- F is the total force required to move the object upward
Work to move object
The work required to move the object a certain distance is:
W = Fx
where
- W is the work in newton-meters (N-m) or foot-pounds (ft-lbs)
- x is the distance moved in meters (m) or feet (ft)
Work is sum of factors
Seeing that the total force is the sum of the force to overcome gravity and the force to overcome inertia, the equation for the work required to move the object upward is:
W = Wg + Wi
W = Fgx + Fix
W = mgx + max
This is the sum of the work to overcome gravity and the work to overcome inertia.
Work against gravity and inertia
Work when ending at zero velocity
When you lift a stationary object, you typically accelerate it to some velocity and then lift it near the desired height, at which time you slow your lifting effort until the object has zero velocity. A common example is lifting an object off the floor to place on a table.
Thus, the work done happens in three steps:
- Doing work against gravity and inertia until you reach a given velocity
- Doing work only against gravity by lifting at a constant velocity
- Doing work against gravity and providing negative work against inertia in slowing the velocity to zero
From zero to a given velocity
The work done in moving an object from zero to some given velocity is:
W1 = mgx1 + max1
where
- W1 is the work against gravity and inertia for the first step in the process
- x1 is the distance moved until the object reaches a given velocity
Moving at constant velocity
The work done in moving an object at constant velocity is:
W2 = mgx2
where
- W2 is the work against gravity for the second step in the process
- x2 is the distance moved until near the desired height
Moving from some velocity to zero
The work done in moving an object from some given velocity to zero is:
W3 = mgx3 − max3
where
- W3 is the work against gravity with negative work against inertia for the third step in the process
- x3 is the distance moved until the object reaches a zero velocity
What this means is that the force you are applying is less than the force required to lift the object, because you are overcoming its inertia in reducing its velocity to zero.
Total work
The total work is:
W = W1 + W2 + W3
W = mgx1 + max1 + mgx2 + mgx3 − max3
To simplify things, let's assume that the distances for acceleration and deceleration are the same. Thus, the total work is:
W = mgx1 + mgx2 + mgx3
W = mgx
where x is the total desired height.
Note that although that this equation or W = mgh is given in most Physics textbooks, there seldom is mentioned the fact that the object must be accelerated and decelerated to reach the desired height. That fact is important for understanding of the principles involved.
The work done when the final velocity is zero is independent of the work against inertia, because negative work cancels out the positive work.
Examples
The following examples illustrate the two situations in work against gravity and inertia.
Shot putter lifts a lead weight
Before throwing 16-pound lead ball, a shot putter warms up his muscles by slowly lifting it from his shoulder to the maximum height of his arm, 2.5 feet away. How must work does he do?
Answer
Since he is lifting the ball from zero velocity to zero velocity, he is only doing work against gravity.
The weight of the ball is 16 lbs, so its mass is 16/32 or 0.5 pounds-mass. The work against gravity is:
Wg = mgx
Wg = (0.5 pound-mass)*(32 ft/s2)*(2.5 ft)
Wg = 40 foot-pounds
Shot putter throws the ball
Then the shot putter throws the 16-pound lead weight straight up in the air. In accelerating the weight, his hand moves 2.5 feet in 1 second until the ball leaves his hand. How must work did he do now?
Answer
We know that the work against gravity is:
Wg = 40 foot-pounds
Now, consider the acceleration of the ball to find the work to overcome inertia. The weight moved from zero to an average of 2.5 ft/s in 1 second. That meant its final speed was 5 ft/s and its acceleration was 5 ft/s2.
The weight of the ball is 16 lbs, so its mass is 16/32 or 0.5 pounds-mass. Thus, the work done against inertia is:
Wi = max
Wi = (0.5 pound-mass)*(5 ft/s2)*(2.5 ft)
Wi = 6.25 foot-pounds
The total work is:
W = Wg + Wi
W = (40 + 6.25) foot-pounds
W = 46.25 foot-pounds
You can see that the he must do extra work against inertia in this situation.
Summary
When an object is thrown upward, you must overcome both inertia and gravity. Assuming a constant accelerating force, the work done is W = mgx + max.
When the object is simply lifted to a height, at which point its velocity is zero, the work done is W = mgx.
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Resources
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Gravity and Gravitation Resources
Books
Top-rated
books on Simple Gravity Science
Top-rated
books on Advanced Gravity Physics
Mini-quiz to check your understanding
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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