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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
Question
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. What is the change in kinetic energy of the fancart over this 1.5 second interval?
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Physics
3 years
2021-08-13T17:35:04+00:00
2021-08-13T17:35:04+00:00 1 Answers
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Answers ( )
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
Change in Kinetic energy = final kinetic energy – initial kinetic energy
ΔKE = KE (final) – KE(initial)
ΔKE![Rendered by QuickLaTeX.com = \frac{1}{2} m(v^2-u^2) = \frac{1}{2}* 0.8(0.05^2-0.8^2) = -0.255 \ J](https://documen.tv/wp-content/ql-cache/quicklatex.com-d025d27e268559ba4626d3949920f76f_l3.png)
Therefore, the change in kinetic energy of the fancart over 1.5 second time interval is -0.255 J