Published by:
CGP EDU Academic Team
Published on: September 13, 2026
The figure shows the force (F) versus displacement (s) graph for a particle of mass m = 2 kg initially at rest.

(i) the maximum speed of the particle occurs at x = ...................
(ii) the maximum speed of the particle is ......................
(iii) the particle once again has its speed zero at x = .....................
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: Analyze the area under the force vs. displacement graph.
The total work done on the particle can be calculated using the area under the curve of the F-s graph.
Step 2: Calculate maximum speed occurrence.
The maximum speed occurs when the total work done is maximized, which is at the peak of displacement where the force first reduces to zero.
Step 3: Use the work-energy principle.
Work done = K.E. = \frac{1}{2}mv^2.
With rising speed until the zero-force point indicates maximum kinetic energy at that point.
Step 4: Compute maximum speed with the work done obtained from the area. Assuming the area gives work = 4 Joules, then setting up the equation:
\frac{1}{2}mv^2 = Work
4 J = 1 kg v^2
v^2 = 4 => v = 2 m/s.
Step 5: Evaluate further speed returns to zero at the end of displacement where force becomes negative (if applicable).
Therefore, (i) Max speed at x = peak displacement, (ii) Max speed = 2 m/s, (iii) Speed returns to zero at final turn in direction or peak retraction of graph.
The total work done on the particle can be calculated using the area under the curve of the F-s graph.
Step 2: Calculate maximum speed occurrence.
The maximum speed occurs when the total work done is maximized, which is at the peak of displacement where the force first reduces to zero.
Step 3: Use the work-energy principle.
Work done = K.E. = \frac{1}{2}mv^2.
With rising speed until the zero-force point indicates maximum kinetic energy at that point.
Step 4: Compute maximum speed with the work done obtained from the area. Assuming the area gives work = 4 Joules, then setting up the equation:
\frac{1}{2}mv^2 = Work
4 J = 1 kg v^2
v^2 = 4 => v = 2 m/s.
Step 5: Evaluate further speed returns to zero at the end of displacement where force becomes negative (if applicable).
Therefore, (i) Max speed at x = peak displacement, (ii) Max speed = 2 m/s, (iii) Speed returns to zero at final turn in direction or peak retraction of graph.
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