Work |({W})| is a force acting on an object, causing it to travel a distance. This results in a transfer of energy.
There are two essential conditions for this transfer to occur.
- A force must be applied to the object receiving the energy. At least one component of the force must be in the direction of the motion.
- The object receiving the energy must be moved.
If a ball travelling at a constant velocity is not subject to friction forces, is it doing any work?
Even if the ball is travelling at a distance, it cannot be said to be doing any work, as there are no forces acting on it.
If a ball travelling at a constant velocity is not subject to friction forces, is it doing any work?
Even if the ball is travelling at a distance, it cannot be said to be doing any work, as there are no forces acting on it.
There is a mathematical relationship between work, force and the distance travelled by the object.
The equation used to calculate work is:
|W = F \times \triangle x|
where
|W| represents the work |\small (\text {J})|
|F| represents the force |\small (\text {N})|
|\triangle x| represents the distance travelled by the object |\small (\text {m})|
The answer obtained is a scalar quantity, which means that the work has no direction.
The above formula can be used whenever the force and distance travelled are parallel.
A box is pushed on a distance of |\small 12 \: \text {m}|. If the friction force is|\small 25 \: \text {N}|, what was the work done by the friction ?
In this situation, the friction force is parallel to the distance travelled. However, it is in the opposite direction, because the friction force is a force that opposes the distance travelled by an object. The force will therefore be negative, since it is directed in the opposite direction to the distance travelled.
||\begin{align} W = F \times \triangle x
\quad \Rightarrow \quad
W &= - 25 \: \text {N} \times 12 \: \text {m} \\
&= -300 \: \text {J} \end{align}||
The friction has therefore done a work of |-300 \: \text {J}|. Negative work represents a loss of energy. The friction force has therefore subtracted energy from the box.
When the force and the distance travelled are not parallel to each other, the component of the force that is parallel to the distance travelled must be determined.
The formula to use to calculate the work when the force and distance travelled are not parallel is:
|W = F \times \cos \theta \times \triangle x|
where
|W| represents the work |\small (\text {J})|
|F \times \cos \theta| represents the force component parallel to the distance travelled |\small (\text {N})|
|\triangle x| represents the distance travelled by the object |\small (\text {m})|
We apply a force of |\small 150 \: \text {N}| on a sled over a distance of |\small 200 \: \text {m}| as shown in the illustration below. What amount of work has been involved on the sled ?
The information provided in this problem is as follows.
||\begin{align} F &= 150 \: \text {N} &\triangle x &= 200 \: \text {m}\\
\theta &= 30^{\circ} &W &=\: ? \end{align}||
Using the previous formula, it is possible to determine the work
||\begin{align} W = F \times \cos \theta \times \triangle x
\quad \Rightarrow \quad
W&= 150 \: \text {N} \times \cos 30^{\circ} \times 200 \: \text {m} \\
&= 25\:981 \: \text {J} \end{align}||
The work done on the sled is approximately |25\:981 \: \text {J}|.
If several forces are applied to an object, it is best to find the resultant force and then determine the work done on the object.
Some editors use the variable |\triangle S| to represent the distance travelled. This variable is the same as the variable |\triangle x| used in the formula written above. However, the use of |\triangle x| is preferred, as it more easily represents horizontal motion, i.e a motion along the x-axis.
Mechanical power |\small (P)| is the ratio between the amount of work done and the time required to do it.
The greater the amount of work transferred per second, the greater the mechanical power.
The following formula should be used to calculate the power:
|P = \displaystyle \frac {W}{\triangle t}|
where
|P| represents the mechanical power |\small (\text {W})|
|W| represents the work |\small (\text {J})|
|\triangle t| represents the variation of time |\small (\text {s})|
A worker manages to lift a motor using the work of |\small 2 \: 000 \: \text {J}| for |\small 5 \: \text {s}|. How much power did the worker supply ?
||\begin{align} W &= 2 \: 000 \: \text {J} &\triangle t &= 5 \: \text {s}\\
P &= \: ? \end{align}||
||\begin{align} P = \frac {W}{\triangle t}
\quad \Rightarrow \quad
P &= \frac {2\: 000 \: \text {J}}{5 \: \text {s}} \\
&= 400 \: \text {W} \end{align}||
Useful work and work done are two concepts used in the study of simple machines. In fact, the purpose of the simple machine is to receive energy from the user and redirect it to the object that is to be travelled. The primary purpose of the simple machine is to reduce the force to be applied to an object.
The work done |(W_{f})| is defined as the energy that the user transmits to the simple machine.
To determine the work done, the driving force supplied by the user must be multiplied by the distance travelled by this force.
The equation used to calculate the work done is:
|W_{f} = F_{m} \times \triangle x_{m}|
où
|W_{f}| represents the work done |\small (\text {J})|
|F_{m}| represents the driving force |\small (\text {N})|
|\triangle x_{m}| represents the distance travelled by the object |\small \text {(m)}|
The useful work |(W_{u})| is defined as the energy received by the object being moved.
To determine the useful work, we need to multiply the resistive force required to move the object by the distance travelled.
The equation used to calculate the useful work is:
|W_{u} = F_{r} \times \triangle x_{r}|
where
|W_{u}| represents the useful work |\small (\text {J})|
|F_{r}| represents the resistance force |\small (\text {N})|
|\triangle x_{r}| represents the (resisting) distance travelled by the object |\small \text {(m)}|
The hoist (a set of pulleys) shown below is used to lift a motor.
The force with which man pulls is the driving force |\small (F_{m})|.
The distance travelled by the person exerting the force (length of rope pulled) is the distance travelled caused by motor force. |(\small \triangle x_{m})|.
The energy that the person transmits to the hoist by pulling on the rope is the work done. |\small (W_{f})|.
The force exerted by the hoist (represented by the weight of the motor) is the resistive force. |\small (F_{r})|.
The distance over which the motor is lifted by the hoist is the resistor distance |\small (\triangle x_{r})|.
The energy that the hoist gives to the motor by lifting it into the air will be the useful energy. |\small (W_{u})|.
In theory, useful work |\small (W_{u})| is always equal to the work done |\small (W_{f})|.On the other hand, in everyday life, in any energy transfer, friction has to be taken into account. So energy losses are to be expected. The sheet on efficiency gives the necessary details on energy losses.
A worker exerts a force of |\small 200 \text { N}| to lift the motor shown on the hoist above. If he lifts this motor |\small 2,5 \text { m}|, determine the useful work and the work done in this situation.
The work supplied is the work done by the worker. We know that he exerts a force of |\small 200 \text { N}|, but we don't know how many metres of rope he has to pull to lift the motor off of |\small 2,5 \text { m}|. The mechanical advantage of this hoist is 4, since 4 strands touch the movable pulleys. As a result, the worker will have to force over a distance 4 times greater than |\small 2,5 \text { m}| in order to lift the motor.
|4 \times 2,5 \text { m} = 10 \text { m}|
To determine the work done:
||\begin{align} F_r &= 200 \: \text {N} & x_r &= 10 \: \text {m}\\
W &=\: ? \end{align}||
||\begin{align} W_{f} = F_{m} \times \triangle x_{m}
\quad \Rightarrow \quad
W_f&= 200 \: \text {N} \times 10 \: \text {m} \\
&= 2 \:000 \: \text {J} \end{align}||
As the worker pulls in the same direction as the rope travelled, the angle between the two vectors is necessarily |\small 0^{\circ}|.
The useful work will be the work done by the hoist on the motor. We know that the hoist lifts the motor by |\small 2,5 \text { m}|, but we don't know the force with which he lifts it. Since the mechanical advantage of the hoist is 4, we know that the hoist exerts a force 4 times greater than the worker.
|4 \times 200 \text { N} = 800 \text { N}|
To determine the useful work:
||\begin{align} F_m &= 800 \: \text {N} & x_m &= 2,5 \: \text {m}\\
W &=\: ? \end{align}||
||\begin{align} W_{u} = F_{r} \times \triangle x_{r}
\quad \Rightarrow \quad
W_u&= 800 \: \text {N} \times 2,5 \: \text {m} \\
&= 2 \:000 \: \text {J} \end{align}||
As the hoist exerts its force in the same direction as the distance travelled by the motor, the angle between the two vectors is necessarily |\small 0^{\circ}|. Moreover, we can see that the work done and the useful work are equal.
The power supplied |(P_{f})| is the speed with which energy is transmitted from the operator to a simple machine.
To calculate the power supplied, use the following formula:
|P_{f} = \displaystyle \frac {W_{f}}{\triangle t}|
où
|P_{f}| represents the power supplied |\small \text {(W)}|
|W_{f}| represents the work done |\small \text {(J)}|
|\triangle t| represents the time |\small \text {(s)}|
Useful power |(P_{u})| is the speed with which energy is transmitted from the simple machine to the object being travelled.
To calculate the useful power, use the following formula:
|P_{u} = \displaystyle \frac {W_{u}}{\triangle t}|
où
|P_{u}| represents the useful power |\small \text {(W)}|
|W_{u}| represents the useful work |\small \text {(J)}|
|\triangle t| represents the time |\small \text {(s)}|