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