Rankers Physics

Work Done by Constant and Variable Forces: Practice Problem & Solution

An object of mass m is tied to string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases (it is pulled extremely slowly so that equilibrium exists at all times) until the string makes an angle θ with the vertical. Work done by the force F is : Work Done by Constant and Variable Forces diagram: An object of mass m is tied to
mgL (1 – cosθ)
mgL sin θ
mgL
FL(1 + tanθ)

Solution Explained:

To solve this problem, we apply the core principles of Work Done by Constant and Variable Forces. Understanding the underlying formula is key to arriving at the correct answer below:

Be definition of Work, Work Done = F.S. cos θ

Taking S and cos θ together like

W= F. (S cos θ)= Force × Component of displacement along force

Here Force F is acting horizontally and displacement toward right is L sin θ, So

Work done = F.L. sin θ

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Work done by a force is always same in all inertial frame of references. Reason (R): Work is an invariant physical quantity.  
2. A uniform force of \(F = (3\hat{i} + \hat{j})\text{ N})\) acts on a particle of mass \(2\text{ kg})\). Hence the particle is displaced from position \(\vec{r_1} = (2\hat{i} + \hat{k})\text{ m})\) to position \(\vec{r_2} = (4\hat{i} + 3\hat{j} - \hat{k})\text{ m})\). The work done by the force on the particle is: (2013)
3. A force of (3ˆi + 4ˆj)N acts on a body and displaced it by (3ˆi + 4ˆj)m . The work done by the force is
4. Assertion (A): There is no term like instantaneous work similar to instantaneous velocity. Reason (R): For work to be done, the force must act for a displacement.  
5. A block of mass $10\text{ kg}$ moving in $x$ direction with a constant speed of $10\text{ m s}^{-1}$, is subjected to a retarding force $F = -0.1x\text{ J/m}$ during its travel from $x = 20\text{ m}$ to $30\text{ m}$. Its final K.E. will be: (2015)