A-particle-of-mass-4kg-was-at-rest-a-a-point-of-position-vector-i-4j-A-force-F-was-applied-to-it-and-it-moved-at-a-velocity-of-3i-7j-ms-1-after-a-time-of-5seconds-Find-a-the-magnitude-o

Question Number 47675 by Rio Michael last updated on 13/Nov/18

A p a r t i c l e o f m a s s 4 k g w a s a t r e s t a a p o i n t o f p o s i t i o n v e c t o r

i + 4 j . A f o r c e F w a s a p p l i e d t o i t a n d i t m o v e d a t a v e l o c i t y

o f (3 i + 7 j) {m s}^{- 1} a f t e r a t i m e o f 5 s e c o n d s . F i n d

a) t h e m a g n i t u d e o f F

b) T h e s p e e d a t w h i c h i t m o v e s, H e n c e,

c) T h e d i s t a n c e i t c o v e r e d .

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

a) f o r c e = m (\frac{{\vec{v}}_{2} - {\vec{v}}_{1}}{t}) = 4 \times (\frac{3 i + 7 j}{5})

m a g n d t u d e o f f o r c e = \sqrt{{(\frac{12}{5})}^{2} + {(\frac{28}{5})}^{2}} = 6.09 N

c) d i s t a n c e s

s = \frac{1}{2} {a t}^{2} = \frac{1}{2} \times \frac{6.09}{4} \times 5^{2} = 19.03

b) a v e r a g e s p e e d = \frac{19.03}{5} = 3.81 m / s e c

Answered by bshahid010@gmail.com last updated on 13/Nov/18

Use F = m \frac{Δ v}{Δ t}

where Δ v = \sqrt{3^{2} + 7^{2}} = \sqrt{58}

F = 4 \frac{\sqrt{58}}{5}

then find a by a = \frac{F}{m}

and use second equation of motion

Commented by Rio Michael last updated on 13/Nov/18

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