Understanding Work Done by a Constant Force

Part A = A. Positive

Part B = C. Zero

Part C = B. Negative

Part D = C. Zero

Part E = W = 0J

Part F = W = wdcos(90 – φ)

Part G = W = 0

Part H = D. None of these

Solution Below:

The word “work” has many meanings when used in everyday life. However, in physics work has a very specific definition. This definition is important to learn and understand. Work and energy are two of the most fundamental and important concepts you will learn in your study of physics. Energy cannot be created or destroyed; it can only be transformed from one form to another. How this energy is transferred affects our daily lives from microscopic processes, such as protein synthesis, to macroscopic processes, such as the expansion of the universe!

When energy is transferred either to or away from an object by a force F acting over a displacement d, work W is done on that object. The amount of work done by a constant force can be found using the equation

W=Fdcos(θ)

where F is the magnitude of the force vector, d is the magnitude of the distance vector, and θ is the angle between the force and distance vectors.

The SI unit for work is the joule, J. A single joule of work is not very big. Your heart uses about 0.5J each time it beats, and the 60-watt lightbulb in your desk lamp uses 216,000J each hour. A joule is defined as follows:

1J = 1N*m = 1(kg*m^{2})/s^{2}

The net work done on an object is the sum of the work done by each individual force acting on that object. In other words,

W_{net} = W_{1} + W_{2} + W_{3} + … W_{i}

The net work can also be expressed as the work done by the net force acting on an object, which can be represented by the following equation:

W_{net} = F_{net}dcos(θ)

Knowing the sign of the work done on an object is a crucial element to understanding work. Positive work indicates that an object has been acted on by a force that tranfers energy to the object, thereby increasing the object’s energy. Negative work indicates that an object has been acted on by a force that has reduced the energy of the object.

The next few questions will ask you to determine the sign of the work done by the various forces acting on a box that is being pushed across a rough floor. As illustrated in the figure (Intro 1 figure), the box is being acted on by a normal force n_vec, the force of gravity (i.e., the box’s weight w), the force of kinetic friction f_{k}, and the pushing force F. The displacement of the box is d.

Intro 1 Figure

Part A

A. Positive

B. Negative

C. Zero

A. Positive

Part B

A. Positive

B. Negative

C. Zero

C. Zero

Part C

A. Positive

B. Negative

C. Zero

B. Negative

Part D

A. Positive

B. Negative

C. Zero

C. Zero

Part E

Since the dresser doesn’t move, d = 0 and:

W = fd

W = 3,500N * 0m

W = 0

Part F

We can use sin(φ)*mg but since Mastering Physics doesn’t give this as an option, we have to use cosin. sin(φ) is equivalent to cosin(90 – φ) and Mastering Physics gives that as an option. Also note that Mastering Physics gives ‘w’ for weight. This is equivalent to mg, so we can make that substitution also. The equation is therefore:

W = wdcosin(90 – φ)

Part G

_{e}, the mass of the Sun is given by M

_{s}, and the Earth-Sun distance is given by r

_{es}.

An object undergoing uniform circular motion experiences a net force that is directed in toward the center of the circle; this net force is called the centripetal force. This force is always perpendicular to the distance the object moves and therefore never does any work on the object. Therefore:

W = 0

Part H

A. W = kx^{2}

B. W = -kx^{2}

C. W = 0

D. None of these

The equation for work presented in this problem requires that the force be constant. Because the force exerted on an object varies with the spring’s displacement from equilibrium F = kx) you cannot use W = Fdcos(θ) to find the work done by a spring. In actuality the work done by a spring is given by the equation

W_{spring} = -1/2kx^{2}

So the answer is:

D. None of these