Selina Solutions Class 9 Physics Solutions for Exercise 3C in Chapter 3 - Chapter 3 Laws Of Motion

Question 27 Exercise 3C

A force acts for 10s on a stationary body of mass 100kg after which the force ceases to act. The body moves through a distance of 100m in the next 5s. Calculate: (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force.


Given: m=100kg, initial velocity u =0, distance covered s=100m;

(i) Velocity of the body= distance covered/time

= 100/5 = 20m/s

(ii) Acceleration can be found using the equation of motion,

\begin{array}{l} \mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as} \\ \mathrm{a}=\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right) / 2 \mathrm{s} \\ =\left(20^{2}-0^{2}\right) / 2 \mathrm{x} 100 \\ \mathrm{a}=2 \mathrm{m} / \mathrm{s}^{2} \end{array}

(iii) Force = ma = 100 x 2 = 200N

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