Answer:
The change in kinetic energy of the fancart over 1.5 second time interval is -0.255 J
Explanation:
Given;
mass of fancart, m = 0.8 kg
initial velocity of fancart, u = < 0.8, 0, 0 > m/s
force exerted by air on fancart, F = < −0.4, 0, 0 > N
duration of the force, t = 1.5 seconds
The change in kinetic energy of the fancart over 1.5 second time interval;
Determine the final velocity, v
F = ma
a = F/m
Apply equation of motion;
v = u + at
[tex]v = u + \frac{F}{m}t[/tex]
[tex]v = 0.8 + \frac{-0.4*1.5}{0.8}\\\\v = 0.8 -0.75\\\\v = 0.05\ m/s[/tex]
Change in Kinetic energy = final kinetic energy - initial kinetic energy
ΔKE = KE (final) - KE(initial)
ΔKE [tex]= \frac{1}{2} m(v^2-u^2) = \frac{1}{2}* 0.8(0.05^2-0.8^2) = -0.255 \ J[/tex]
Therefore, the change in kinetic energy of the fancart over 1.5 second time interval is -0.255 J
The change in kinetic energy of the fancart over 1.5 second time interval is -0.255 J
Since A fancart of mass 0.8 kg initially has a velocity of < 0.8, 0, 0 > m/s. Then the fan is turned on, and the air exerts a constant force of < −0.4, 0, 0 > N on the cart for 1.5 seconds.
v = u + at
v = u + {F}/{m}t
v = 0.8 + {-0.4 * 1.5}/ {0.8}
v = 0.8 - 0.75
= 0.05 m/s
Now
We know that
Change in Kinetic energy = final kinetic energy - initial kinetic energy
So,
= 0.5m(v^2 - u^2)
= 0.5 * 0.8(0.05^2 - 0.8^2)
= -0.255 J
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