Question

A force of 2000 dyne acts upon a body ofmass 1 kg. As a result, thevelocity of the body changes from 20 cm's to 5 cm/s in passing through acertain distance. Find that distance.

Answer 1

$F=2000dyne =0.02N,$

$mass= 1kg$

$u= 20cms^-1 = 0.2ms^-1$

$v=5cms^-1 = 0.05ms^-1$

$\huge\green{\boxed{\mathfrak{SOLUTION}}}$

$F=ma$

$0.02=1×a$

$a=0.02ms^-2$

Using law of motion,

$v^2=u^2+2as$

$(0.05)^2=(0.2)^2+2×0.02×s$

$0.0025=0.04+0.04×s$

$0.0025-0.04=0.04×s$

$-0.0375=0.04×s$

$s=-0.9375$

$s=0.9375$

$\large\green{\underline{\orange{\underline{\purple{\mathfrak{MARK\:AS\:BRAINLIEST}}}}}}$

Answer 2

s=0.9375

have a good day