A body of mass 10 kg. Initially moving with a velocity 4m/s pushed
by a horizontal force of 5 N on a smooth horizontal surface. The work
done by the force in 5 s is
Answers
131.25
As we know s=ut+at²/2
Therefore s=20+25/4
And W=Fs
Therefore W=5[105/4]=131.25
A body of mass 10 kg. Initially moving with a velocity 4m/s pushed
by a horizontal force of 5 N on a smooth horizontal surface. The work
done by the force in 5 s is 131.25
Mass of a body = 10 Kg
Initial velocity = 4m/s
Applied Force = 5N
Initial kinetic energy = (1/2)mu²
Final velocity = v
The acceleration due to 5N force , a = 5/10 = 0.5 m/s²
t = 5 s
Distance Moved ,s = ut + (1/2)at²
= (4*5) + (1/2) * 0.5 * 5 * 5
= 26.25 m
Final velocity
V = u + at
= 4 + 0.5 * 5
= 6.5 m/s
Kinetic energy = (1/2) m v²
Work done by 5N force = Increase in kinetic energy
= (1/2) m v² - (1/2) m u²
= 131.25 J