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# Work and energy (part 2)

More on work. Introduction to Kinetic and Potential Energies. Created by Sal Khan.

## Want to join the conversation?

• I didn't understand what Sal said at . If he is applying the same amount of force as force due to gravity and both forces are equal and opposite in direction
why is the object moving at all? Can someone explain? • Hello Maha, Brilliant Quesiton, i was once in your shoes.

Here is my simple explanation. remember that F=ma. a stands for acceleration. not velocity. So as low as the object is moving with constant velocity. it means that acceleration equals to 0 (because there is no change in velocity) and so the net force is also equal to zero.

take these two examples. You are sitting in a chair, your weight force is equal to reaction force and thus you don't fall through the ground. happy? good.

second example is you go sky diving (great experience, i would highly recommend it), and you accelerate due to gravity. but after some time you reach terminal velocity. that is when the weight force equals to the air friction force. so net force = 0. But you don't stop moving. you just move with constant velocity.

These ideas link with newton's laws. Inertia is a important concept you need to get down.

whenever you get confused with physics in general, i would look at the equations and try to understand and visualize what are they saying. it helped me quite a bit

i hope this helps, best of luck
• what is actually the diference between speed and velocity?
(1 vote) • If potential energy is the ability to do work (), what work does the object do when it falls? Isn't this work (falling of the object) done by gravity? • if the force pulling upwards on the elevator is equal to the force pulling downwards , wouldnt that mean that the elevator wouldnt move? since there are two forces of equal proportion acting on it , from different sides. • At , if F=ma, and the elevator isn't accelerating, why doesn't F=0 and thus work=0? I get, intuitively, why there is work being done, but based on the equations why does it not work out like that? • I know g= 9.8 m/s or rounded to 10 m/s. When do you use either number? Because in this video, in the first example 9.8 is used but in the second example 10 is used. I'm not sure what constituted the use of either numbers in the specific problems.
(1 vote) • so,potential energy only to upwards?how about downwards,is there a potential energy?
(1 vote) • K.E = (mv^2)/2

final velocity = v
initial velocity = u
time = t
mass = m
acceleration = a

But,when I do this:

Work = F * displacement = K.E
⇒ K.E = F * vt ...[∵ displacement = velocity *time]
⇒ K.E = ma * vt ...[∵ F=ma]
⇒ K.E = m((v-u)/t) * vt ...[∵ mass = (v-u)/t]
⇒ K.E = m(v - u) * v ...[the t's cancel out]
⇒ K.E = m(v) * v ...[∵ the body is at rest, initial
velocity is zero and therefore
v-u = v]

Therefore, K.E = mv^2
But,Kinetic energy(K.E) is supposed to be equal to:
K.E = (mv^2)/2

What is wrong with my derivation?Why don't I get :
K.E = (mv^2)/2

Please answer, cause if you don't, all the effort I put into typing this would go to a waste.   