Question

100 kg body is resting on horizontal plane having coefficient of 0.2 a force of 196 newton is required to pull the body horizontally find the acceleration developed by the body ​

Answer 1

Answer:

-0.04

Explanation:

$f = ma \\ \\ f_{r} = umg \\ \\ euation \: \: of \: its \: motion \: will \: be \\ \\ f - umg = ma \\ u = 0.2 \\ m = 100kg \\ f = 196 \: newton \\ g = 10ms {}^{ - 2} \\ a = ? \\ \\ 196 - 0.2 \times 100 \times 10 = 100a \\ a = \frac{ - 4}{100} \\ a = - 0.04ms {}^{ - 2}$

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Answer 2

f=ma

f

r

=umg

euationofitsmotionwillbe

f−umg=ma

u=0.2

m=100kg

f=196newton

g=10ms

−2

a=?

196−0.2×100×10=100a

a=

100

−4

a=−0.04ms

−2

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