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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 \((x_f)\).

Illustration Questions

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.

Illustration Questions

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.

image
A

–1.5 J

.

B

–5 J

C

–6 J

D

–2 J

Option A 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.

 

Illustration Questions

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.

image

Work done is given as 

W = Area of shaded region

W= Area of triangle

\(W=\dfrac {1}{2}×4×20\)

\(W=40\,J\)

image

Calculate the work done from the given force v/s displacement graph.

image
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.

Illustration Questions

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.

image

Work done is given as 

W = Area of shaded region

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

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

\(W=20+20-10\)

\(W=30\,J\)

image

Calculate the work done from the given force v/s displacement graph.

image
A

60 J

.

B

30 J

C

15 J

D

14 J

Option B 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 \((x_f)\).

Illustration Questions

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.

Illustration Questions

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.

image
A

–3 J

.

B

–1.5 J

C

–2 J

D

–4 J

Option B is Correct

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