The F-x graph of a particle of mass 100 gm is shown in figure. If the particle begins to move from rest at x=0, its velocity at x=12 m is

Question

The F-x graph of a particle of mass 100 gm is shown in figure. If the particle begins to move from rest at x=0, its velocity at x=12 m is (A) 10 ms -1 (B) 20 ms -1 (C) 30 ms -1 (D) 40 ms -1

Tardigrade Admin · Accepted Answer

Work done by the force = Area under the

F - x

graph.
From

x = 0

to

12 m,

area under graph

= (\frac{1}{2} \times 4 \times 10) + (4 \times 10) + (\frac{1}{2} \times 4 \times 10) = 80 J

W = Δ k

As particle starts from rest,

W = 80 J = \frac{1}{2} m v^{2} - 0

\frac{1}{2} (0.1 k g) v^{2} = 80 J

\Rightarrow v^{2} = \frac{2 \times 80}{0.1} = 1600

or

v = 40 m s^{- 1}

.

Q. The $F - x$ graph of a particle of mass $100 g m$ is shown in figure. If the particle begins to move from rest at $x = 0,$ its velocity at $x = 12 m$ is

$10 m s^{- 1}$

$20 m s^{- 1}$

$30 m s^{- 1}$

$40 m s^{- 1}$

Q. The F−x graph of a particle of mass 100gm is shown in figure. If the particle begins to move from rest at x=0, its velocity at x=12m is

10ms−1

20ms−1

30ms−1

40ms−1

Work done by the force = Area under the F−x graph. From x=0 to 12m, area under graph =(21​×4×10)+(4×10)+(21​×4×10)=80J W=Δk As particle starts from rest, W=80J=21​mv2−0 21​(0.1kg)v2=80J ⇒v2=0.12×80​=1600 or v=40ms−1.

Q. The $F - x$ graph of a particle of mass $100 g m$ is shown in figure. If the particle begins to move from rest at $x = 0,$ its velocity at $x = 12 m$ is

$10 m s^{- 1}$

$20 m s^{- 1}$

$30 m s^{- 1}$

$40 m s^{- 1}$