1.

A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. `450 J`B. `275 J`C. `250 J`D. `475 J`

Answer» `F = ma`
`a = (F)/(m) = (0.1 x)/(10) = 0.01 x`
`v (dvb)/(dx) = - 0.01 x`
`int_(10)^(v) v dv = - 0.01 int_(20)^(30) x dx`
`|(v^(2))/(2)|_(10)^(v) = - 0.01 |(x^(2))/(2)|_(20)^(30)`
`v^(2) - 100 = - 0.01 [(30)^(2) - (20)^(2)] = - 5`
`v^(2) = 95`
Final `K.E. K_(f) = (1)/(2) mv^(2) = (1)/(2) xx 10 xx 95 = 475 J`


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