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?

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

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

a and c

c is 100% but i am confused in a

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j