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 A body of mass m=10−2 kg is moving in a medium and experiences a frictional force F=−kv2. Its initial speed is v0= 10 ms−1. If, after 10 s, its energy is ⅛ mv02, the value of k will be 

(a) 10−3 kg m−1

(b) 10−3 kg s−1

(c) 10−4 kg m−1

(d) 10−1 kg m−1 s−1

in Physics by Expert (10.9k points) | 19 views

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 (c) 10−4 kg m−1 is the correct answer 

Mass m = 102 Kg

Initial velocity v0= 10 ms−1

Time t = 10 s

Frictional Force F=− kv2

1/2 mvf= ⅛ mv02

v= v/2

v= 10 /2 = 5 m/s

Now, the force is

(10-2)dv/dt = -kv

\(\int ^5_{10} {dv\over v^2}=-100k\int^{10}_0 dt\) 

\({1\over5}={1\over10}=100k\times 10\) 

\(k=10^{-4} \space {kg \over m}\)

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