It works by you putting leverage on one side makes more force go to the other side so if you put a crowbar in between a door and you push on one side the other will push the opposite side with more force<span />
Answer:
No, it is not conserved
Explanation:
Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.
The total kinetic energy before the collision is:
where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.
After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is
So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.
Answer:
the work done by him is 2,340 joules
Explanation:
The computation of the work done by him is shown below:
= Weight of man × height
= 450 N × 5.2 meters
= 2,340 joules
Hence, the work done by him is 2,340 joules
We simply multiply the man weight with the height
The speed of the vehicle would be sqrt 250 000 J / 0.5 * 100 kg =~70.71m/s. However this only is the speed when ALL energy is converted into heat, and that is absurd. The temperature of the vehicle (which I suggest is mostly made out of steel) should rise with 250 kJ / 100 kg * 0.49 kJ/kgK = 5.1 K, which is quite a lot for just braking. So how much should the speed be for normal energy conversion from the total amount of energy? We don't know the speed the vehicle slows down, and even if we knew it would be hard to calculate. So it is quite unrealistic, but the number you are looking for is approximatly 70.71 m/s.