# 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

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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1. minhkhang

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

$$v = u + \frac{F}{m}t$$

$$v = 0.8 + \frac{-0.4*1.5}{0.8}\\\\v = 0.8 -0.75\\\\v = 0.05\ m/s$$

Change in Kinetic energy = final kinetic energy – initial kinetic energy

ΔKE = KE (final) – KE(initial)

ΔKE $$= \frac{1}{2} m(v^2-u^2) = \frac{1}{2}* 0.8(0.05^2-0.8^2) = -0.255 \ J$$

Therefore, the change in kinetic energy of the fancart over 1.5 second time interval is -0.255 J