A body of mass 2 kg fall vertically- passing through two points A and B- The speeds of the body as it passes A and B are 1 m-s and 4 m-s respectively- The resistance against which the body falls is 9-6 N- What is the distance AB-2 m3 m1-5 m6 m

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Toppr Admin · Accepted Answer

U-1msv-4msa-Fma-9-62-4-8ms2usingv2-u2-2as-xA0-42-12-2-xD7-4-8-xD7-S-xA0-S-1-5625m

A body of mass $2 k g$ fall vertically, passing through two points $A$ and $B$ . The speeds of the body as it passes $A$ and $B$ are $1 m / s$ and $4 m / s$ respectively. The resistance against which the body falls is $9.6 N$ . What is the distance $A B$ ?
$2 m$
$3 m$
$1.5 m$
$6 m$

$u = 1 \frac{m}{s}$

$v = 4 \frac{m}{s}$

$a = \frac{F}{m}$

$a = \frac{9.6}{2} = 4.8 \frac{m}{s^{2}}$
using

$v^{2} = u^{2} + 2 a s$

$4^{2} = 1^{2} + 2 \times 4.8 \times S$

$S = 1.5625 m$

A body of mass 2 kg fall vertically, passing through two points A and B. The speeds of the body as it passes A and B are 1 m/s and 4 m/s respectively. The resistance against which the body falls is 9.6 N. What is the distance AB?2 m3 m1.5 m6 m

u=1msv=4msa=Fma=9.62=4.8ms2usingv2=u2+2as 42=12+2×4.8×S S=1.5625m

A body of mass $2 k g$ fall vertically, passing through two points $A$ and $B$ . The speeds of the body as it passes $A$ and $B$ are $1 m / s$ and $4 m / s$ respectively. The resistance against which the body falls is $9.6 N$ . What is the distance $A B$ ?
$2 m$
$3 m$
$1.5 m$
$6 m$

$u = 1 \frac{m}{s}$

$v = 4 \frac{m}{s}$

$a = \frac{F}{m}$

$a = \frac{9.6}{2} = 4.8 \frac{m}{s^{2}}$
using

$v^{2} = u^{2} + 2 a s$

$4^{2} = 1^{2} + 2 \times 4.8 \times S$

$S = 1.5625 m$