Informative line

Work Done By Constant Force

Learn formula for work done by constant force, gravity and kinetic friction. Practice to calculate of work done by gravity on path & kinetic friction and constant force.

Constant Force

  • A constant force is a force which does not depend upon the parameters like time \((t)\), distance \((x)\), etc.

OR

A force which is constant in magnitude as well as in direction is called constant force.

For Example : Motion of a particle under force of gravity for small distance.

 

Illustration Questions

Which of the following is a constant force?

A \(\dfrac{kq}{x^2}\)

B \(10\,N\)

C \(k\,x\)

D \(k\,x^2\)

×

Options \((A),\;(C)\) and \((D)\) are incorrect because all these forces depend upon distance \((x)\). Thus, these are not constant forces.

Option \((B)\) is correct because the force is constant in magnitude as well as in direction. Thus, it is a constant force.

Which of the following is a constant force?

A

\(\dfrac{kq}{x^2}\)

.

B

\(10\,N\)

C

\(k\,x\)

D

\(k\,x^2\)

Option B is Correct

Work Done by Constant Force

  • To calculate the work done by a constant force, we use following definition of work -

\(W=\vec F\;.\vec s\)

\(\Rightarrow\;W=Fs\,cos\,\theta\)

where \(\theta\) is the angle between the force vector and displacement vector.

Illustration Questions

Calculate the work done by a constant force \(\vec F=10\,\hat i\,N\) at an angle \(\theta=60°\), when a body is displaced by \(s=20\,m\).

A \(90\,J\)

B \(30\,J\)

C \(100\,J\)

D \(20\,J\)

×

Work done by a constant force is given by

\(W=Fs\,cos\,\theta\)

where \(\theta\) is the angle between the force vector and displacement vector.

Given: \(\vec F=10\,\hat i\,N,\,s=20\,m,\,\theta=60°\)

\(W=(10)(20)\,cos\,60°\)

\(W=(10)(20)\left(\dfrac{1}{2}\right)\)

\(W=100\,J\)

Calculate the work done by a constant force \(\vec F=10\,\hat i\,N\) at an angle \(\theta=60°\), when a body is displaced by \(s=20\,m\).

image
A

\(90\,J\)

.

B

\(30\,J\)

C

\(100\,J\)

D

\(20\,J\)

Option C is Correct

Work by Constant Force is Path Independent

  • Work done by constant force is given by

\(W=Fs\,cos\,\theta\)

\(\Rightarrow\;W=F(s\,cos\,\theta)\)

\(\Rightarrow\;W=F\) (component of displacement along force vector)

  • From the figure, it is clear that work done by a constant force is the product of force and displacement in the direction of force.
  • Thus, the component of displacement perpendicular to the force does not play any role in work done by a constant force, even if it is large.

  • Now, consider that work is done by a constant force to move a particle from \(A\) to \(B\), via two paths.
  • The initial and final position of particle are \(A\) and \(B\) respectively, as shown in figure.
  • The path is broken into two components, one is parallel to force and other is perpendicular to force.
  • As we have already proved that the component perpendicular to the force does not play any role in work done by constant force.
  • By adding all the components parallel to the force, the distance \(x\) always comes equal, whichever be the path.
  • Thus, the work done by a constant force is path independent.

Illustration Questions

Which case holds true for the work done on a particle by a constant force via different paths, as shown in figure?

A \(W_1>W_2\)

B \(W_2>W_1\)

C \(W_1=W_2\)

D

×

The path is broken into two components, one is parallel to force and other is perpendicular to force.

image

Since, the component perpendicular to the force does not play any role in work done by constant force.

image

Therefore, by adding all the components parallel to the force, the distance \(x\) always comes equal, whichever be the path.

Thus, the work done by a constant force is path independent.

image

Which case holds true for the work done on a particle by a constant force via different paths, as shown in figure?

image
A

\(W_1>W_2\)

.

B

\(W_2>W_1\)

C

\(W_1=W_2\)

D

Option C is Correct

Calculation of Work Done by Constant Force

  • Work done by constant force is given by

\(W=Fs\,cos\,\theta\)

\(\Rightarrow\;W=F(s\,cos\,\theta)\)

\(\Rightarrow\;W=F\) (component of displacement along force vector)

  • From above diagram, it is clear that work done by a constant force is the product of force and displacement in the direction of force.
  • Thus, the component of displacement perpendicular to the force does not play any role in work done by a constant force, even if it is large.

Illustration Questions

A constant force is applied on two objects along \(x\)-axis. One object moves from origin to \(P\) and other moves from origin to \(Q\) under the applied force. The coordinates of \(P\;\&\;Q\) are given. In which one of the following combinations, the work done by constant force is equal?

A \(P(3,\,4)\;\&\;Q(3,\,6)\)

B \(P(3,\,4)\;\&\;Q(4,\,3)\)

C \(P(3,\,4)\;\&\;Q(6,\,4)\)

D \(P(3,\,3)\;\&\;Q(4,\,4)\)

×

Since, the constant force is applied along \(x\)-axis and the work done is independent of the component perpendicular to force i.e., along \(y\)-axis.

Thus, the work done by constant force is equal for the combination having same \(x\)-coordinates.

Hence, option \((A)\) is correct.

A constant force is applied on two objects along \(x\)-axis. One object moves from origin to \(P\) and other moves from origin to \(Q\) under the applied force. The coordinates of \(P\;\&\;Q\) are given. In which one of the following combinations, the work done by constant force is equal?

A

\(P(3,\,4)\;\&\;Q(3,\,6)\)

.

B

\(P(3,\,4)\;\&\;Q(4,\,3)\)

C

\(P(3,\,4)\;\&\;Q(6,\,4)\)

D

\(P(3,\,3)\;\&\;Q(4,\,4)\)

Option A is Correct

Independence of Work Done by Gravity on Path

  • Force of gravity can be assumed as a constant force. It always acts downwards i.e., towards the center of the earth.

Work Done by Gravity 

The work done by gravity is \((mg)\) times the displacement in the direction of gravity.

Mathematically,

\(W=mg\,(\Delta y)\)

where \(\Delta y\) is the displacement in the direction of gravity.

  • If the displacement in the direction of gravity i.e., \(\Delta y\) is same for two bodies then work done by gravity is same for two bodies.

Illustration Questions

A ball of mass \(m=2\,kg\) is dropped from some height \(\Delta y=5\,m\). Calculate the work done by gravity on the ball.  [Given: \(g=10\,m/s^2\)]

A \(190\,J\)

B \(100\,J\)

C \(180\,J\)

D \(170\,J\)

×

The work done by gravity is given by

\(W=mg\,(\Delta y)\)

where, \(\Delta y\) is the displacement in the direction of gravity.

Given: \(m=2\,kg,\;g=10\,m/s^2,\;\Delta y=5\,m\)

\(W=(2)\,(10)\,(5)\)

\(=100\,J\)

A ball of mass \(m=2\,kg\) is dropped from some height \(\Delta y=5\,m\). Calculate the work done by gravity on the ball.  [Given: \(g=10\,m/s^2\)]

A

\(190\,J\)

.

B

\(100\,J\)

C

\(180\,J\)

D

\(170\,J\)

Option B is Correct

Work Done by Gravity

  • Force of gravity can be assumed as a constant force. It always acts downwards i.e., towards the center of the earth.

Work Done by Gravity 

The work done by gravity is \((mg)\) times the displacement in the direction of gravity.

Mathematically,

\(W=mg\,(\Delta y)\)

where \(\Delta y\) is the displacement in the direction of gravity.

  • If the displacement in the direction of gravity i.e., \(\Delta y\) is same for two bodies then work done by gravity is same for two bodies.

Illustration Questions

Which case holds true for the work done by gravity for the paths shown in figure?

A \(W_1>W_2\)

B \(W_2>W_1\)

C \(W_1=W_2\)

D

×

If the displacement in the direction of gravity i.e., \(\Delta y\) is same for two bodies then work done by gravity is same for two bodies.

Thus, option \((C)\) is correct.

Which case holds true for the work done by gravity for the paths shown in figure?

image
A

\(W_1>W_2\)

.

B

\(W_2>W_1\)

C

\(W_1=W_2\)

D

Option C is Correct

Calculation of Work Done by Kinetic Friction

  • When a constant force is applied on an object along displacement (i.e., along tangent to the path), then the work done by the force is force times distance.
  • Consider a path shown in figure.
  • To calculate the total work done by the constant force, the path is divided into small parts.
  • The total work done will be the sum of all the small work done by all the small parts.

  • If a block is sliding on a surface, then the force of friction always acts opposite to the displacement.
  • Then, the work done will depend upon the path i.e., distance.
  • Work done by frictional force :
  • \(dW=-f.\,ds\)

    \(dW=-\mu _kN.ds\)

     

Illustration Questions

Calculate the total work done by a constant frictional force \(f=10\,N\), when a block is displaced from \(A\) to \(B(s_1=2\,m)\) and then \(B\) to \(C(s_2=1\,m)\). 

A \(-60\,J\)

B \(-30\,J\)

C \(70\,J\)

D \(-20\,J\)

×

Work done by frictional force :

\(W=\) Frictional force × displacement

\(W= -f.s\)

Given : \(f=10\,N,\,s_1=2\,m\) 

Work done from \(A\) to \(B\),

\(W_1=-10×2\)

\(W_1=-20\,J\)

Given : \(f=10\,N,\,s_2=1\,m\)

Work done from \(B\) to \(C\),

\(W_2=-10×1\)

\(W_2=-10\,J\)

Total work done,

\(W=W_1+W_2\)

\(W=-20+(-10)\)

\(W=-30\,J\)

Calculate the total work done by a constant frictional force \(f=10\,N\), when a block is displaced from \(A\) to \(B(s_1=2\,m)\) and then \(B\) to \(C(s_2=1\,m)\). 

image
A

\(-60\,J\)

.

B

\(-30\,J\)

C

\(70\,J\)

D

\(-20\,J\)

Option B is Correct

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