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charger9198
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A horizontal force of 80N acts on a mass of 6 Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant
Showing all working Calculate the total energy expanded in the acceleration;
S=(.5)at^2
5=(.5)a(0.92^2)
=0.4232 a
a=5/0.4232 = 11.8147 ms^2
V=at
=11.8147*0.92
=10.8695 ms^-1
s=(.5)a(t^2)
s=4.999 m
Work done;
W=F*d=80*5= 400J
Power;
W=P*t
P=W/t=400/0.92=434.78 watt
The final velocity of the mass
Vf^2=Vi^2+2ad (vi=0, as the mass is initially at rest)
F=m*a
a=F/m=80/6=13.33 m/s^2
Vf^2=2*13.33*5=133.33
Vf=11.55m/s
the final momentum
p=mv
=6*11.55=69.3 N/s
the final kinetic energy
K=0.5*m*v^2=533 J
Have I answered the fundamental points of the question of is there irrelevant things I could remove, I'm just trying to answer the question as thoroughly and best I can..thanks
Showing all working Calculate the total energy expanded in the acceleration;
S=(.5)at^2
5=(.5)a(0.92^2)
=0.4232 a
a=5/0.4232 = 11.8147 ms^2
V=at
=11.8147*0.92
=10.8695 ms^-1
s=(.5)a(t^2)
s=4.999 m
Work done;
W=F*d=80*5= 400J
Power;
W=P*t
P=W/t=400/0.92=434.78 watt
The final velocity of the mass
Vf^2=Vi^2+2ad (vi=0, as the mass is initially at rest)
F=m*a
a=F/m=80/6=13.33 m/s^2
Vf^2=2*13.33*5=133.33
Vf=11.55m/s
the final momentum
p=mv
=6*11.55=69.3 N/s
the final kinetic energy
K=0.5*m*v^2=533 J
Have I answered the fundamental points of the question of is there irrelevant things I could remove, I'm just trying to answer the question as thoroughly and best I can..thanks