Ncert Exercises & Solutions

Question 5.10:

A body of mass 0.40 kg moving initially with a constant speed of $10 m s^{- 2}$ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

$G i v e n,$

$m = 0.40 k g, u = 10 m s^{- 1} (d u e t o n o r t h)$

$F = - 8 N (f o r c e i n o p p o s i t e d i r e c t i o n)$

$A s F = m a$

$\Rightarrow a = \frac{F}{m} = \frac{- 8}{0.40} = - 20 m s^{- 2} [f o r 0 \leq t \leq 30 s]$

$(i) A t t = - 5$

$x = u t = 10 \times (- 5) = - 50 m$

$(i i) A t t = 25 s$

$x = u t + \frac{1}{2} a t^{2}$

$= 10 \times 25 + \frac{1}{2} (- 20) {(25)}^{2}$

$= - 6000 m$

$(i i i) A t t = 100 s$

$F o r u p t o 30 s$

$x_{1} = u t + \frac{1}{2} a t^{2} = 10 \times 30 + \frac{1}{2} (- 20) {(30)}^{2}$

$= - 8700 m$

$A t t = 30 s$

$v = u + a t = 10 - 20 \times 30 = - 590 m s^{- 1}$

$F o r 30 s t o 100 s$

$x_{2} = v t = - 590 \times 70 = - 41300 m$

$\Rightarrow x = x_{1} + x_{2} = - 8700 - 41300 = - 50000 m$

$= - 50 k m$

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