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### Work Done By Variable Force

Learn how to work done by variable force using graphs & spring formula, practice to calculation of work when force is a function of x.

# Calculation of Work when Force is a Function of x

## Variable Force

It is a force whose magnitude is a function of position /displacement, i.e., $$F=f(x)$$

### Work Done by Variable Force

• Suppose a force is acting on a body which is not constant but depends upon displacement.
• Work done by constant force is expressed as the product of force and displacement, only when force is constant.

$$W=F_{\text {constant}}×\text {Displacement}$$

• Thus, this expression can't be used to calculate work done by variable force.
• To solve this problem, the displacement is divided into very small parts.
• Since, the parts are so small, so we can assume that force during that part of displacement is constant.
• So, summing up the work done by these small displacements, gives the total work.
• Consider an example, a force $$F=k\,x$$ is acting on a body where $$x$$ is the distance from origin.
• As we move from A to B, the value of the force changes (increases).
• So, to calculate the work done by force from A to B, we can't take any value of force.  • To calculate total work for this, we divide AB into as many parts as possible.
• Now, considering an element of displacement of thickness $$dx$$ at a distance $$x$$
• This element is so small that we can assume force at $$x$$ to be $$k\,x$$ and force at $$(x+dx)$$ to be $$k(x+\Delta x)\approx kx$$ as $$\Delta x\to 0$$.  • So, work done by small element $$dx$$ is

$$W=\vec F \cdot d\vec x$$

• Since the displacement and force are opposite, making an angle of 180°

so, $$W=F \,dx\,×cos\,180°$$

$$W=- F \cdot dx$$

• So, the total work done from all parts of displacement from A to B

$$W_ T=-\sum\limits _{\text {From 1st element}}^{\text {To last element}}F.dx$$

• This can also be calculated by using calculus.

$$W_ T=-\int\limits_{\text {From 1st element}}^{\text {To last element}}\;F.dx$$

• In short, we can say that to find total work done by variable force following steps can be followed:
1. Write $$F$$ as a function of $$x$$.
2. Write $$x$$as $$dx$$.
3. Integrate the force from first element $$(x_i)$$ to final element $$(x_f)$$.  #### The force of electric field between two charges is given by $$\dfrac {kq_1q_2}{x^2}$$, where $$x$$ is the distance between them. Calculate the work done by electric force if it is initially at $$r$$ and moves to distance infinity.

A $$W=\dfrac {kq_1q_2}{x^2}\,(r)$$

B $$W=\displaystyle\int\limits_{r}^{\infty}\dfrac {kq_1q_2}{x^2}\,dx$$

C $$W=\displaystyle\int\limits_{\infty}^{x}\dfrac {kq_1q_2}{x^2}\,dx$$

D $$W=\displaystyle\int\limits_{r}^{\infty}{x\,kq_1q_2}\,dx$$

×

The total work done by electric force $$F(x)$$ is given as

$$W_T=\int\limits_{\text {1st element}}^{\text {Last element}}\,F(x)\;dx$$

Since, $$F(x)=\dfrac {kq_1q_2}{r^2}$$ and the charges move from $$r$$ to infinity. So, work done will be

$$W_T=\displaystyle\int\limits_{r}^{\infty}\dfrac {kq_1q_2}{x^2}\,dx$$

Hence, option (B) is correct.

### The force of electric field between two charges is given by $$\dfrac {kq_1q_2}{x^2}$$, where $$x$$ is the distance between them. Calculate the work done by electric force if it is initially at $$r$$ and moves to distance infinity.

A

$$W=\dfrac {kq_1q_2}{x^2}\,(r)$$

.

B

$$W=\displaystyle\int\limits_{r}^{\infty}\dfrac {kq_1q_2}{x^2}\,dx$$

C

$$W=\displaystyle\int\limits_{\infty}^{x}\dfrac {kq_1q_2}{x^2}\,dx$$

D

$$W=\displaystyle\int\limits_{r}^{\infty}{x\,kq_1q_2}\,dx$$

Option B is Correct

# Work Done by Spring

• Spring follows Hook's law which states that force experienced by spring is directly proportional to the extension or compression from its natural length.

$$F\propto x$$

• Spring will exert a force $$K_x$$ where $$K$$ is a constant and $$x$$ is the distance from natural length, as shown in figure.
• The direction of spring force is always along its natural length.
• So, when the spring is stretched or compressed, the force of spring continuously increases as $$x$$ increases.
• For the calculation of work done by spring force, integration is used, as force is variable.

$$W=\displaystyle\int\limits_{x_i}^{x_f}-Kx\;dx=-\dfrac {1}{2}K\Big[ x_f^2-x_i^2\Big]$$

Work done $$=\dfrac {1}{2}Kx_i^2-\dfrac {1}{2}Kx_f^2$$  ## Negative and Positive Work by Spring

• When the spring is stretched or compressed, force and displacement are opposite to each other, thus work done by spring force is negative.
• When a spring is stretched, its distance increases from natural length but the force acts towards the natural length. Thus, both act opposite to each other.

Conclusion: On moving away from natural length, work done by spring force is negative and on moving towards natural length, work done is positive.  #### A spring having natural length $$x$$ and spring constant 100 N/m is displaced from A to B. The initial and final displacement are $$x_i=10\,cm$$ and $$x_f=20\,cm$$. Calculate the work done by spring force.

A –1.5 J

B –5 J

C –6 J

D –2 J

×

Since, the displacement is in direction opposite to force so, work done by spring force is given as

$$W=-\dfrac {1}{2}K\Big[ x_f^2-x_i^2 \Big]$$

where, $$K=$$ constant

$$x_i$$ = initial displacement

$$x_f$$= final displacement

Given: $$K=100\,N/m$$ , $$x_i=10\,cm=0.1\,m$$$$x_f=20\,cm=0.2\,m$$

$$W_T=-\dfrac {100}{2}\Big[ (0.2)^2-(0.1)^2 \Big]$$

$$=-\dfrac {100}{2}\Big[ 0.04-0.01 \Big]$$

$$=-\dfrac {100}{2}×0.03$$

$$W=-1.5\,J$$

### A spring having natural length $$x$$ and spring constant 100 N/m is displaced from A to B. The initial and final displacement are $$x_i=10\,cm$$ and $$x_f=20\,cm$$. Calculate the work done by spring force. A

–1.5 J

.

B

–5 J

C

–6 J

D

–2 J

Option A is Correct

# Work Done by Variable Force

## Variable Force

It is a force whose magnitude is a function of position /displacement, i.e., $$F=f(x)$$

### Work Done by Variable Force

• Suppose a force is acting on a body which is not constant but depends upon displacement.
• Work done by constant force is expressed as the product of force and displacement, only when force is constant.

$$W=F_{\text {constant}}×\text {Displacement}$$

• Thus, this expression can't be used to calculate work done by variable force.
• To solve this problem, the displacement is divided into very small parts.
• Since, the parts are so small, so we can assume that force during that part of displacement is constant.
• So, summing up the work done by these small displacements, gives the total work.
• Consider an example, a force $$F=k\,x$$ is acting  on a body where $$x$$ is the distance from origin.
• As we move from A to B, the value of the force changes (increases).
• So, to calculate the work done by force from A to B, we can't take any value of force.  • To calculate total work for this, we divided AB into as many parts as possible.
• Now, considering on element of displacement of thickness $$dx$$ at a distance $$x$$
• This element is so small that we can assume force at $$x$$to be $$k\,x$$ and force at $$(x+dx)$$ to be $$k(x+\Delta x)\simeq kx$$ as $$\Delta x\to 0$$.  • So, work done by small element $$dx$$ is

$$W=\vec F \cdot d\vec x$$

• Since, the displacement and force are opposite making an angle of 180°

so,   $$W=F \,dx\,×cos\,180°$$

$$W=- F \cdot dx$$

• So, the total work done from all parts of displacement from A to B.

$$W_ T=-\displaystyle\sum\limits_{\text {From 1st element}}^{\text {To last element}}Fdx$$

• This can also be calculated by using calculus.

$$W_ T=-\displaystyle\int\limits_{\text {From 1st element}}^{\text {To last element}}\;F\;dx$$

• In short, we can say that to find total work done by variable force following steps can be followed:
1. Write $$F$$ as a function of $$x$$.
2. Write $$x$$as $$dx$$.
3. Integrate the force from first element $$(x_i)$$ to final element $$(x_f)$$.  #### The force of electric field between two charges is given as $$F=\dfrac {10}{x^2}$$, where $$x$$ is the distance between them. Calculate the work done by electric force if $$x$$ varies from 1 m to 2 m.

A 15 J

B 6 J

C 5 J

D 10 J

×

The total work done by electric force $$F(x)$$ is given as

$$W_T=\displaystyle\int\limits_{\text {From 1st element}}^{\text {To Last element}}\,F(x)\;dx$$

Since, $$F(x)=\dfrac {10}{x^2}$$ and $$x$$varies from 1 m to 2 m

So, work done will be

$$W_T=\displaystyle\int\limits_{1}^{2}\dfrac {10}{x^2}\,dx$$

$$W_T=10\displaystyle\int\limits_{1}^{2}x^{-2}\,dx$$

$$W_T=10 \left [ \dfrac {x^{-2+1}}{-2+1} \right]_1^2$$

$$W_T=10 \left [ \dfrac {-1}{x} \right]_1^2$$

$$W_T=10 \left [ \dfrac {-1}{2}+\dfrac {1}{1} \right]$$

$$W_T=10 \left [ \dfrac {1}{2} \right]$$

$$W_T=5\,J$$

### The force of electric field between two charges is given as $$F=\dfrac {10}{x^2}$$, where $$x$$ is the distance between them. Calculate the work done by electric force if $$x$$ varies from 1 m to 2 m.

A

15 J

.

B

6 J

C

5 J

D

10 J

Option C is Correct

# Work Done when Spring Slides in Vertical Direction

• Spring follows Hook's law which states that force experienced by spring is directly proportional to the extension or compression from its natural length.

$$F\propto x$$

• Spring will exert a force $$Kx$$ where $$K$$ is a constant and $$x$$ is the distance from natural length as shown in figure.
• The direction of spring force is always along its natural length.
• So, when the spring is stretched or compressed, the force of spring continuously increases as $$x$$ increases.
• For the calculation of work done by spring force, integration is used, as force is variable.

$$W=\displaystyle\int\limits_{x_i}^{x_f}-Kx\;dx=-\dfrac {1}{2}K\Big[ x_f^2-x_i^2\Big]$$

Work done $$=\dfrac {1}{2}Kx_i^2-\dfrac {1}{2}Kx_f^2$$  • Consider a spring having natural length (NL) = $$\ell\,$$cm and spring constant $$K$$
• Its one end is attached to a rigid support and another end is attached to a small ring which can slide on a smooth vertical wire, as shown in figure.
• Since, work done by spring force is given as

$$W=-\displaystyle\int\limits_{\text {Initial displacement}(x_i)}^{\text {Final displacement}(x_f)} K\,x\;dx$$

$$=-\dfrac {1}{2}K\Big[ x_f^2-x_i^2 \Big]$$

$$=\dfrac {1}{2}K\,x_i^2-\dfrac {1}{2}K\,x_f^2$$  ### Calculation of  $$x_i$$ and $$x_f$$

• Since when a spring is stretched or compressed, the variation occurs from its natural length $$\ell$$.

• Thus, natural length of spring is considered as reference.

• So, calculation of limits can be done as $$X=|NL-\text {Length}|$$
• So, initial displacement $$(x_i)=$$ | Natural length – Initial length | $$=|\ell - \ell_1|$$ and

final displacement $$(x_f)=$$| Natural length – Final length | $$=|\ell - \ell_2|$$

Conclusion: Work done by a spring force only depends on the initial and final position of spring with respect to its natural length.

#### A spring has its natural length of 3 cm with spring constant of $$10^4\,N/m$$. Its one end is attached to a rigid support and another end is attached on a small ring which can slide over a smooth vertical wire as shown. Find the work done by a spring force when the ring slides from position A to B.

A –3 J

B –1.5 J

C –2 J

D –4 J

×

Work done by spring force is given as

$$W=\dfrac {1}{2}\,Kx_i^2-\dfrac {1}{2}\,Kx_f^2$$

where,

$$x_i=$$ Initial displacement

$$x_f=$$ Final displacement

$$K=$$ Spring constant

Initial displacement $$(x_i)=$$ | Natural length – Initial position|

$$x_i=|3-4|=1\,cm$$ and

Final displacement $$(x_f)=$$|Natural length – Final position|

$$x_f=|3-5|=2\,cm$$  $$\Big[$$$$\therefore$$ From figure final position =$$\sqrt {(4)^2+(3)^2}\Rightarrow OB=5\,cm$$ $$\Big]$$

Thus, work done by spring force

$$W=\dfrac {1}{2}×10^4\,[(10^{-2})^2-(2×10^{-2})^2]$$

$$W=\dfrac {1}{2}×10^4×(-3)×10^{-4}$$

$$W=-1.5\,J$$

### A spring has its natural length of 3 cm with spring constant of $$10^4\,N/m$$. Its one end is attached to a rigid support and another end is attached on a small ring which can slide over a smooth vertical wire as shown. Find the work done by a spring force when the ring slides from position A to B. A

–3 J

.

B

–1.5 J

C

–2 J

D

–4 J

Option B is Correct

# Work Done by Variable Force using Graphs

• Work done by a variable force can be calculated using graphs also.
• The graph of force v/s displacement, provides the work done.
• The area between the graph and x-axis of a force v/s displacement graph gives the work done by force.
• Consider a force $$F=kx$$, the work done can be calculated as

$$W= \displaystyle\int\limits_0^x\,k\,x\;dx$$

$$=k\displaystyle\int\limits_0^x\,x\;dx$$

$$=k\left [ \dfrac {x^2}{2} \right ]_0^x$$

$$= \dfrac {k\,x^2}{2}$$

• Now, using graph the work done will be calculated as

Area of shaded region = work done = $$\dfrac {1}{2}×(x)×(kx)$$

$$=\dfrac {1}{2}×k\,x^2$$

• Thus the work done is same as of calculus.  #### Calculate the work done from the given force v/s displacement graph.

A 20 J

B 50 J

C 80 J

D 40 J

×

The area between the graph and x-axis of a force v/s displacement graph gives the work done by force. Work done is given as

W = Area of shaded region

W= Area of triangle

$$W=\dfrac {1}{2}×4×20$$

$$W=40\,J$$ ### Calculate the work done from the given force v/s displacement graph. A

20 J

.

B

50 J

C

80 J

D

40 J

Option D is Correct

# Negative Work Done using Graph for Area below x-axis

• Work done by a variable force can be calculated using graphs also.
• The graph of force v/s displacement, provides the work done.
• The area between the graph and x-axis of a force v/s displacement graph gives the work done by force.
• Consider a force $$F=-kx$$, the work done can be calculated as

$$W=\displaystyle\int\limits_0^x\,-k\,x\;dx$$

$$=-k\displaystyle\int\limits_0^x\,x\;dx$$

$$=-k\left [ \dfrac {x^2}{2} \right ]_0^x$$

$$= -\dfrac {k\,x^2}{2}$$

• Now, by using graph, work done will be calculated as

Area of shaded region = Work done = $$\dfrac {1}{2}×(x)×(-kx)$$

$$=\dfrac {-1}{2}×k\,x^2$$

• Thus the work done is same as of calculus.  #### Calculate the work done from the given force v/s displacement graph.

A 60 J

B 30 J

C 15 J

D 14 J

×

The area between the graph and x-axis of a force v/s displacement graph gives the work done by force. Work done is given as

W = Area of shaded region

W = Area of triangle OBC + Area of triangle BCD + Area of triangle DEF

$$W=\dfrac {1}{2}×4×10+ \dfrac {1}{2}×4×10+ \dfrac {1}{2}×2×(-10)$$

$$W=20+20-10$$

$$W=30\,J$$ ### Calculate the work done from the given force v/s displacement graph. A

60 J

.

B

30 J

C

15 J

D

14 J

Option B is Correct