Explanation of Work Against Gravity and Inertia by an External Force by Ron Kurtus - Succeed in Understanding Physics. Key words: upward acceleration, velocity, negative work, physical science, School for Champions. Copyright © Restrictions

# Work Against Gravity and Inertia by an External Force

by Ron Kurtus (revised 3 February 2011)

When you lift an object to some height above the ground, you are doing work against the downward force of gravity. The work done is the product of the applied force and the displacement.

If the object is initially stationary, the upward force must first overcome both inertia and gravity to start to move the object. Once the object is moving upward at some velocity, a force only equal to that from gravity is necessary to continue the upward movement at that velocity.

There are two common situations in projecting or lifting an object upward. You can project the 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 to some height.

Questions you may have include:

• What is the work done against gravity and inertia?
• What is the work in projecting an object upward?
• How much work is required to lift an object to some height?

This lesson will answer those questions.

Useful tool: Metric-English Conversion

## Work

Work is the effort over a displacement against a force that is resisting motion or is pulling in the opposite direction that you want to move the object. In other words, work equals the product of the force that overcomes the resistance and the displacement in the same direction as the force:

W = Fd

where

• W is the work done
• F is the force applied against a resistive force
• d is the displacement in the same direction as the force

Although F and d are vector quantities with an indicated direction, W is a scalar quantity, with only magnitude and no direction.

### Work against gravity

The force of gravity resists motion in its opposite direction. If an upward force equal to the force of gravity—or the weight of an object—is applied to a stationary object, the forces equal out, and the object does not move. However, if the object has an initial upward velocity and a force equal to gravity is applied, the object will continue to move upward at that initial velocity.

The upward force is:

−Fg = m(−gu)

Fg = mgu

where

• −Fg is the upward force needed to counter the force of gravity in newtons or pounds-force
• m is the mass of the object in kilograms or pounds-mass
• −gu is the acceleration in the opposite direction of the acceleration due to gravity (−9.8 m/s2 or −32 ft/s2)

Note: According to our convention for direction in gravity equations, Fg and gu are negative numbers, since they are in the opposite direction of gravity.

The work done in moving an object against gravity a certain displacement at an initial upward velocity is:

W = (−Fg)*(−y)

W = Fgy

where

• W is the work done against gravity in joules (J) or foot-pounds-force
• −y is the vertical displacement in meters (m) or feet (ft), measured from the starting point to when the force is discontinued

Note: Our convention states that y is negative when it is in the opposite direction of gravity.

Thus:

W = m(−gu)(−y)

W = mguy

Note: When talking about work against gravity, most physics textbooks use h for height: W = mgh. However, you need to remember that h is the displacement that an object is lifted above the ground, while y is the displacement from some starting point at or above the ground.

### Work against inertia

Inertia is a resistance to changing the motion of an object. Its equation is:

−Fi = m(−au)

Fi = mau

where

• −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
• −au is the upward acceleration of the object in m/s2 or ft/s2

The work against inertia in accelerating an object a displacement upward is:

W = Fiy

W = mauy

Note: At this point, we are not considering accelerating the object against the force of gravity. This equation is for the general work against inertia.

## Work in projecting an object upward

When you project an object upward, you must not only overcome the force of gravity, but you must also overcome the resistance of inertia.

The force to project an object upward is a sum of the force needed to overcome gravity and the force required to overcome the object's inertia:

−F = −Fg − Fi

F = Fg + Fi

### Work

If you apply those forces over the displacement, the amount of work is:

W = Wg + Wi

W = mguy + mauy

Work against gravity and inertia

### Object continues upward

However, once you stop applying the accelerating force to overcome inertia, the object will continue to move upward, due to its momentum. The velocity at the point where the force was stopped will be the initial velocity for an object projected upward.

(See Overview of Gravity Equations for Objects Projected Upward for more information.)

### Example: Shot putter throws a lead ball

Suppose a shot putter throws the 16-pound lead ball 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 much work must he do to throw the ball?

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 = mguy

Wg = (0.5 pound-mass)*(−32 ft/s2)*(−2.5 ft)

Wg = 40 foot-pounds-force

Now, consider the acceleration of the ball to find the work to overcome inertia. Since the ball traveled 2.5 feet in 1 second, its average speed is 2.5 ft/s. However, since the ball is starting at 0 ft/s, the end speed must be 5 ft/s to have an average of 2.5 ft/s. Thus, the acceleration is 5 ft/s2.

The mass of the ball 0.5 pounds-mass, and the work done against inertia is:

Wi = mauy

Wi = (0.5 pounds-mass)*(−5 ft/s2)*(−2.5 ft)

Wi = 6.25 foot-pounds-force

The total work is:

W = Wg + Wi

W = (40 + 6.25) foot-pounds

W = 46.25 foot-pounds-force

You can see that the he must do extra work against inertia in this situation.

## 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:

1. First, you do work against gravity and inertia until you reach a given velocity.

2. Then, you do work only against gravity alone, by lifting at a constant velocity

3. Finally, you do work against gravity but provide negative work against inertia in slowing the velocity to back to zero.

### From zero to a given velocity

The work done in moving an object from zero to some given velocity is:

W1 = mguy1 + mauy1

where

• W1 is the work against gravity and inertia for the first step in the process
• y1 is the vertical displacement moved until the object reaches a given velocity

### Moving at constant velocity

The work done in moving an object at constant velocity is:

W2 = mguy2

where

• W2 is the work against gravity for the second step in the process
• y2 is the vertical displacement 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 = mguy3 − mauy3

where

• W3 is the work against gravity with negative work against inertia for the third step in the process
• y3 is the vertical displacement 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 = mguy1 + mauy1 + mguy2 + mguy3 − mauy3

To simplify things, let's assume that the displacements for acceleration and deceleration are the same. Thus, the total work is:

W = mguy1 + mguy2 + mguy3

W = mguy

where y is the total desired height or displacement.

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.

### Example: Putting a heavy box on a shelf

You want to lift a 34 kg-force (75 pound-force) weight up to a shelf that is 2 meters off the ground. You accelerate the box from being stationary to a velocity of 2 m/s, lift it up and then decelerate the box back to 0 m/s when you place it on the shelf. How much work must you do?

Since you are lifting the box from zero velocity to zero velocity, the work against inertia cancels out, and you are only doing work against gravity.

The weight of the ball is 34 kg-force, so its mass is 34/9.8 or 3.47 kg-mass. The work against gravity is:

Wg = mguy

Wg = (3.47 kg-mass)*(−9.8 m/s2)*(−2 m)

Wg = 68 joules

## Summary

Work against gravity and inertia is achieved by applying a sufficient external force to accelerate an object a certain displacement in the opposite direction of gravity. The work is the product of the applied force and the displacement.

When an object is thrown upward, you must overcome both gravity and inertia. Assuming a constant accelerating force, the work done is W = mguy + mauy. When the object is simply lifted to a height, at which point its velocity is zero, the work done is W = mguy.

Hard work will get you far in life

## Resources and references

Author's Credentials

### Websites

Work by gravity by Sunil Kumar Singh - Connexions

Gravity and Inertia in Running - Locomotion and Biology paper (PDF)

Gravity Resources

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