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.

- 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.

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

B \(10\,N\)

C \(k\,x\)

D \(k\,x^2\)

- 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.

A \(90\,J\)

B \(30\,J\)

C \(100\,J\)

D \(20\,J\)

- 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.

A \(W_1>W_2\)

B \(W_2>W_1\)

C \(W_1=W_2\)

D

- 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.

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

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

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

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

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

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.

A \(190\,J\)

B \(100\,J\)

C \(180\,J\)

D \(170\,J\)

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

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.

A \(W_1>W_2\)

B \(W_2>W_1\)

C \(W_1=W_2\)

D

- 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 :
- \(dW=-f.\,ds\)
\(dW=-\mu _kN.ds\)

A \(-60\,J\)

B \(-30\,J\)

C \(70\,J\)

D \(-20\,J\)