A variable force F starts acting on a block of mass 5 kg resting on a smooth horizontal surface

Name: Problem 22, Chapter 4: A Complete Resource Book in Physics for JEE Main
Uploaded: 2021-10-06T13:34:14-08:00
Duration: 2 min 29 s
Channel: Dheeraj Sharma
Description: A variable force F starts acting on a block of mass 5 kg resting on a smooth horizontal surface. F is varying with displacement x as shown in F-x curve. The velocity of body when its displacement is 3 m will be (A) 2 ms^-1 (B) 2 √(2) ms^-1 (C) 2 √(3) ms^-1 (D) 6 ms^-1

step 1 In this question we have given a variable force that starts acting on a block of mass 5 KG which

Question

A variable force $F$ starts acting on a block of mass $5 \mathrm{~kg}$ resting on a smooth horizontal surface. $F$ is varying with displacement $x$ as shown in $F-x$ curve. The velocity of body when its displacement is $3 \mathrm{~m}$ will be (A) $2 \mathrm{~ms}^{-1}$ (B) $2 \sqrt{2} \mathrm{~ms}^{-1}$ (C) $2 \sqrt{3} \mathrm{~ms}^{-1}$ (D) $6 \mathrm{~ms}^{-1}$

   A variable force $F$ starts acting on a block of mass $5 \mathrm{~kg}$ resting on a smooth horizontal surface. $F$ is varying with displacement $x$ as shown in $F-x$ curve. The velocity of body when its displacement is $3 \mathrm{~m}$ will be
(A) $2 \mathrm{~ms}^{-1}$
(B) $2 \sqrt{2} \mathrm{~ms}^{-1}$
(C) $2 \sqrt{3} \mathrm{~ms}^{-1}$
(D) $6 \mathrm{~ms}^{-1}$

A Complete Resource Book in Physics for JEE Main

Sanjeev Kumar 1st Edition

Chapter 4, Problem 22

Instant Answer

Solved by Expert Dheeraj Sharma

10/06/2021

Step 1

The work done by the force is equal to the area under the force-displacement graph. The area under the graph is the sum of the area of the rectangle and the area of the triangle. Show more…

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A variable force $F$ starts acting on a block of mass $5 \mathrm{~kg}$ resting on a smooth horizontal surface. $F$ is varying with displacement $x$ as shown in $F-x$ curve. The velocity of body when its displacement is $3 \mathrm{~m}$ will be (A) $2 \mathrm{~ms}^{-1}$ (B) $2 \sqrt{2} \mathrm{~ms}^{-1}$ (C) $2 \sqrt{3} \mathrm{~ms}^{-1}$ (D) $6 \mathrm{~ms}^{-1}$

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Key Concepts

Work Done by a Variable Force

This concept refers to the calculation of work as the integration of force over a displacement when the force is not constant. Instead of simply multiplying force by displacement, one must consider how the force changes over the path, often by finding the area under the force-displacement curve.

Work-Energy Theorem

The work-energy theorem states that the total work done on an object is equal to the change in its kinetic energy. This principle enables us to determine the velocity of an object by equating the work calculated from the force-displacement relationship to the kinetic energy acquired by the object.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion, and it is given by the formula KE = 0.5 mv². Calculating the kinetic energy (using the work done) allows us to solve for the velocity of the object after it has been acted on by a force.

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