What is Work?
Applying force on an object causes that object to move in the direction of the force. This movement in relation to the force is defined as work. You can complete work using a constant force or a variable force.
What is a Constant Force?
Work done by a force can be divided into work from a constant force and work from a variable force. In the former kind, the magnitude and direction of the force remain unaltered. In such a case, work (W) is equal to the force applied (F) multiplied by displacement (\[\Delta\]x). Therefore,
W = F x \[\Delta\]x
What is Variable Force?
Work done by the variable force is a bit more complex. In such a case, the magnitude and direction of force can change at any time during the work. Most of the work that we complete in our daily life is an example of variable force work. Calculating the same is quite complex and requires integration.
To form a better understanding of the same, let us consider the workings of a spring.
Revising Hooke’s Law
Hooke’s Law states that the spring force for a compressed or stretched spring is equal in magnitude to the force for extension or compression of the spring. However, this spring force has an opposite direction to this extension.
The figure above relates the force on the spring vs displacement when displacement is 0 for an unstretched spring. Us is the elastic potential energy for a stretched spring. This force-displacement graph for spring can help in assessing force according to Hooke’s Law.
Fs = -kx
Thus, Ws = Fs x vdt
Integration and Formula for Variable Force Work
For work done by a variable force, however, you need to apply integration to arrive at accurate results. Therefore,
\[W_{s} = \int_{0}^{t}F_s .vdt\]
\[W_{s} = \int_{0}^{t} - kx v_{x} dt\]
\[W_{s} = \int_{x^0}^{x} - kx dx\]
\[W_{s} = -\frac{1}{2} k\Delta x^{2}\]
Consequently, by using this approach mentioned above, one can easily derive the work done by variable force.
Quick Exercise – 1
Mass (m) of an object is 2kg. This object undergoes variable force in direction ‘x’. Force variation is a function, Fx = (3 + .2x) N. Determine the work done when object moves from x = 0 to x = 5.
Solution
Work done \[ = \int_{x^a}^{x^b} fdx\]
\[W = \int_{0}^{5} (3+.2x) dx\]
W = 17.5 J
Quick Exercise – 2
A bullet weighing 20g is moving at a velocity of 500m/s. This bullet strikes a windowpane and passes through it. Now, its velocity is 400m/s. Calculate work done by a bullet when passing through this obstacle.
Solution
We need to determine the change in kinetic energy in this equation. You know that kinetic energy change \[\Delta\] (K.E.) =\[\frac{1}{2} * \left ( mv_{1}^{2} - mv _{2}^{2}\right)\]
Therefore, in this equation, m is 20g or 0.02 kg. Initial velocity (v1) is 500m/s, while final velocity (v2) is 400m/s.
Thus, \[\Delta\] (K.E.) = \[\frac{1}{2} * {0.02 (500)^{2} - 0.02 (400)^{2}}\]
\[\Delta\] (K.E.) = 900 Joules.
Deriving Work Done by a Constant Force with Integration
Just as you can derive the work for a variable force using calculus, you can do the same for work done by a constant force. In such a calculation, pressure remains unchanged, which is why you can take it out of the equation immediately.
After doing this, you will arrive at an equation, where
Work = \[P\int_{a}^{b} dv\].
\[W = P\Delta V\]
As you can see, the product is the same that we would have evaluated from considering force and distance. Thus, following this integration method for work done by a constant force is redundant.
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This article provides an explanation of work done by a variable force, its formula, and derivations. The frequently asked questions at the end of this article can help you with your doubts if you encounter any while understanding the same.
Variable Force
When a force acts upon a body changes with time(varies) it is known as variable force. Work done by a variable force is calculated by Integration.
Formula
Integration can be used for calculating both, that is, work done by a variable force and work done by a constant force.
\[W_{s} = \int_{0}^{t}Fs .vdt\]
\[W_{s} = \int_{0}^{t} - kx v_{x} dt\]
\[W_{s} = \int_{x0}^{x} - kx dx\]
\[W_{s} = -\frac{1}{2} k\Delta x^{2}\]
Derivation
A force does work when it results in movement.
Work done by a constant force of magnitude F on a point moves a displacement
\[ \Delta x\]x in the direction of the force is the product
\[W=F ⋅ \Delta x\]
In the case of a variable force, integration is essential to calculate the work done. For example, let us consider work done by a spring. Hooke’s law states that the spring or restoring force of a perfectly elastic spring is proportional to its extension but opposite to the direction of extension.
All the small contributions to the work done during small time intervals, for a variable force.
FAQs on Work Done By A Variable Force
1. What is the fundamental difference in how work is calculated for a constant force versus a variable force?
The key difference lies in the calculation method. For a constant force, work is simply the dot product of the force and the total displacement (W = F ⋅ d). However, for a variable force, the force changes with position. Therefore, we must calculate the work over infinitesimally small displacements and sum them up using integration. The formula becomes W = ∫ F(x) dx, where the work is the integral of the force with respect to displacement.
2. What is a variable force and what are some real-world examples?
A variable force is a force whose magnitude, direction, or both, change as an object moves. Most forces encountered in daily life are variable. Key examples include:
- Spring Force: The force exerted by a spring changes with its extension or compression, as described by Hooke's Law (F = -kx).
- Gravitational Force: The gravitational force between two objects, like a satellite and the Earth, varies with the distance separating them.
- Electrostatic Force: The force between two charges changes with the distance between them, according to Coulomb's Law.
3. How is the work done by a variable force represented and calculated on a Force-Displacement graph?
On a graph plotting Force versus Displacement, the work done by a variable force is represented by the area under the curve between the initial and final positions. This graphical area is the geometric representation of the integral ∫ F(x) dx. By calculating this area, you can determine the total work done on the object without performing the integration manually.
4. What are the two essential conditions for work to be done on an object?
For work to be done on an object in a physical sense, two conditions must be met:
1. An external force must be applied to the object.
2. The object must experience a displacement, and there must be a component of the force in the direction of this displacement. If the force and displacement are perpendicular, no work is done. The SI unit for work is the Joule (J).
5. Why is integration the necessary mathematical tool for finding the work done by a variable force?
Since a variable force changes continuously with position, we cannot use simple multiplication (Force × Displacement). Integration is required because it allows us to break down the total displacement into an infinite number of infinitesimally small segments (dx). Over each tiny segment, the force F(x) can be considered almost constant. We find the small work (dW = F(x)dx) for each segment and then use integration to sum up all these small pieces of work to get the total work done over the entire path.
6. How does the Work-Energy Theorem apply when the work is done by a variable force?
The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy (W_net = ΔK). This fundamental principle holds true regardless of whether the force is constant or variable. When the net force is variable, the work (W_net) is calculated using the integral ∫ F_net(x) dx. The result of this integral will still be precisely equal to the object's change in kinetic energy (½mv_f² - ½mv_i²).
7. Can the work done by a variable force be negative or zero? Explain with an example.
Yes, the work done by a variable force can be negative or zero.
- Negative Work: Work is negative when the force component is opposite to the direction of displacement. For example, when you stretch a spring, your pulling force does positive work, but the spring's restoring force pulls inward while the displacement is outward, so the work done by the spring is negative.
- Zero Work: Work can be zero if the net displacement is zero. If a block attached to a spring is stretched and then allowed to return to its original equilibrium position, the total work done by the spring force over this round trip is zero.
8. What is a common misconception when dealing with work done by a spring force?
A common misconception is assuming the work done by a spring is calculated using the force at the final position (F = kx) multiplied by the total displacement (x). This is incorrect because the spring force is not constant; it starts at zero and increases linearly to kx. The correct approach is to use the average force (½ kx) or, more formally, to integrate the force function, F(x) = kx, which gives the correct formula for work done in stretching a spring: W = ½ kx².
