A body of mass 10 kg moving with a velocity of 25 m/s is stopped by a uniform force in a distance of 50 m. Calculate the force.
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Answers
Answer:
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we have u = 15m/s, v = 0, t = 3s and we want to find acceleration, a, so we use equation (3) v = u + at
0 = 15 + 3a so a = -5m/s^2 (the negative meaning deceleration)
Then from F = ma,
F = 10kg(-5m/s^2) = -50N
So the required force is 50N in the direction opposing the initial velocity.
Answer:
Given:-
Mass = 10kg
Initial velocity (u) = 25m/s
Distance (S) = 50m
To find:-
Force
By applying, v^2-u^2 = 2as
0 - 625 = 2×a×50
625 = 100a
a = 625/100
= 6.25m/s^2.
By applying, F = ma
= 10×6.25
= 62.5N.