Question

plz help me






Find distance traveled by a body before coming to rest , if it is moving with a Speed of 10ms^-1 and coefficient of friction between the ground and body is 0.4​

Answer 1

Answer:

The friction force is Fp = μFn where Fn is the normal force, Fn = mg in this case.

So Fp = 0.45mg but in general, F = ma so 0.45mg = ma or a = 0.45g

So a = -0.45(9.81) = -4.4145 (a is negative as a retarding acceleration)

From v^2 = u^2 + 2as, we find 0 = 15^2 – 2(4.4145)s

So s = 225/8.829 = 25.48m

The body slides for 25.48m before coming to rest.

Explanation:

have a great day .

Answer 2

Answer:

sorry no answer for this question

Explanation:

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