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Example 2.9. A force acting on a particle of mass {1}{2} {~kg}

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Example 2.9. A force acting on a particle of mass $\frac{1}{2} kg$ , displaces it through $10 m$ in the first second. By how much will it displace the parlicle in the next second? Sol. We have $S = 10 m$ for $t = 1 s$ and $u = 0$ using $S = u t + \frac{1}{2} a t^{2}$ $10 = 0 + \frac{1}{2} a \times 1$ or $a = 20 m / s^{2}$ Let the displacement in the first two second be $S_{1}$ . Then, $S_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 20 \times (2)^{2} = \frac{1}{2} \times 20 \times 4 = 40 m$

Name: Video solution 1: Example 2.9. A force acting on a particle of mass 21 kg, displaces it through 10 m in the first second. By how much will it displace the parlicle in the next second? Sol. We have S=10 m for t=1s and u=0 using S=ut+21at2 10=0+21a×1 or a=20 m/s2 Let the displacement in the first two second be S1. Then, S1=ut+21at2=0+21×20×(2)2=21×20×4=40 m
Uploaded: 2023-02-08T14:51:22.000Z
Duration: 4 min
Description: Video solution 1: Example 2.9. A force acting on a particle of mass 21 kg, displaces it through 10 m in the first second. By how much will it displace the parlicle in the next second? Sol. We have S=10 m for t=1s and u=0 using S=ut+21at2 10=0+21a×1 or a=20 m/s2 Let the displacement in the first two second be S1. Then, S1=ut+21at2=0+21×20×(2)2=21×20×4=40 m

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Updated on: Feb 8, 2023

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Harper Discussed

Example 2.9. A force acting on a particle of mass

\frac{1}{2} kg

, displaces it through

10 m

in the first second. By how much will it displace the parlicle in the next second? Sol. We have

S = 10 m

for

t = 1 s

and

u = 0

using

S = u t + \frac{1}{2} a t^{2}

10 = 0 + \frac{1}{2} a \times 1

a = 20 m / s^{2}

Let the displacement in the first two second be

S_{1}

. Then,

S_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 20 \times (2)^{2} = \frac{1}{2} \times 20 \times 4 = 40 m

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Question Text	Example 2.9. A force acting on a particle of mass $\frac{1}{2} kg$ , displaces it through $10 m$ in the first second. By how much will it displace the parlicle in the next second? Sol. We have $S = 10 m$ for $t = 1 s$ and $u = 0$ using $S = u t + \frac{1}{2} a t^{2}$ $10 = 0 + \frac{1}{2} a \times 1$ or $a = 20 m / s^{2}$ Let the displacement in the first two second be $S_{1}$ . Then, $S_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 20 \times (2)^{2} = \frac{1}{2} \times 20 \times 4 = 40 m$
Updated On	Feb 8, 2023
Topic	Work, Energy and Power
Subject	Physics
Class	Class 12
Answer Type	Video solution: 1
Upvotes	81
Avg. Video Duration	4 min

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