Question

A force 20N acts for 40seconds on a body of mass 5 kg from rest a. Its initial momentum is _____________. b. Its final velocity is _____________. Now the force is withdrawn, d. Its displacement in the next 20 seconds is _____________. e. Its total displacement is ________________. f. Its average velocity is _________________.

Answer 1

Answer:

a. 0 ; b. 160 m/s ; d. 3200m ; e. 6400m ; f. 106.67 m

Explanation:

Given :

F= 20 N ; t = 40 s ; m = 5 kg ; initial velocity (u) = 0

a. Initial Momentum (P) = m × u = 5 × 0 = 0

We know  that, F = ma (a - acceleration of body due to applied force)

⇒ 20 N = 5 kg × a

or a = 20/5 = 4 m/$s^2$

From the 1st equation of linear  motion we have-

b. v= u + at

⇒  v = 0 + 4 × 40= 160 m/s

Now the force is withdrawn, therefore a= 0

For uniform motion-

d. Displacement = velocity × time = 160 × 20= 3200 m

e. Displacement is 1st 40 s-

s = ut + $\frac{1}{2} at^2$

s= 1/2 × 4 × 40 × 40 = 3200m

Therefore, total displacement = 3200 + 3200 = 6400 m

f. Average Velocity = Total displacement/total time taken

= 6400/60= 106.67 m/s

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