ICSE Class 9 Physics Question 4 of 20

Laws of Motion — Question 7

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Question

Question 7

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

Answer

(i) Velocity (v) = distance (S)time (t)\dfrac{\text {distance (S)}}{\text {time (t)}}

Given,

s = 100 m in 5 s

Substituting the values in the formula above we get,

v=1005v=20 m s1\text v = \dfrac{100}{5} \\[0.5em] \text v = 20\ \text {m s}^{-1}\\[0.5em]

Hence, velocity acquired by the body = 20 m s-1

(ii) As we know,

v2 - u2 = 2as

Given,

s = 100 m

u = 0

v = 20 m s-1

Substituting the values in the formula above we get,

20202=2×a×100400=200×aa=2 m s220^2 - 0^2 = 2 \times \text a \times 100 \\[0.5em] 400 = 200 \times \text a \\[0.5em] \Rightarrow \text a = 2\ \text {m s}^{-2} \\[0.5em]

Hence, a = 2 m s-2

(iii) As we know,

Force (f) = mass (m) x acceleration (a)

Given,

m = 100 kg

a = 2 m s -2

Substituting the values in the formula above we get,

F=100×2F=200 N\text F = 100 \times 2 \\[0.5em] \Rightarrow \text F = 200\ \text N \\[0.5em]

Hence,

Magnitude of force = 200 N