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

he kicked it 10 feet straight up into the air? i think this works.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.