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

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions