Download Conservation of Energy: Potential and Kinetic Energy and more Lecture notes Classical Physics in PDF only on Docsity! PHYSICS –PHY101 VU © Copyright Virtual University of Pakistan 19 Summary of Lecture 8 – CONSERVATION OF ENERGY 1. Potential energy is, as the word suggests, the energy “locked up” up somewhere and which can do work. Potential energy kam karnay ki salahiat hai! Potential energy can be converted into kinetic energy, 212 vm . As I showed you earlier, this follows directly from Newton’s Laws. 2. If you lift a stone of mass from the ground up a distance , you have to do work against gravity. The (constant) force is , and so . By conservation of energy, the work done by you m x mg W mgx= was transformed into gravitational potential energy whose values is exactly equal to . Where is the energy stored? Answer: it is stored neither in the mass or in the earth - it is stored in mgx the gravitational field of the combined system of stone+earth. 3. Suppose you pull on a spring and stretch it by an amount away from its normal (equilibrium) position. How much energy is sto x red in the spring? Obviously, the spring gets harder and harder to pull as it becomes longer. When it is extended by length and you pull it a further distance , the small amount of work dx dx 2 0 0 one is . Adding up all the small bits of work gives the total work: 1 2 This is the work you did. Maybe you got tired working so x x dW Fdx kxdx W Fdx k xdx k x = = = = =∫ ∫ 21 2 hard. What was the result of your working so hard? Answer: this work was transformed into energy stored in the spring. The spring contains energy exactly equal to . 4. Kinetic energy obvi k x ously depends on the frame you choose to measure it in. If you are running with a ball, it has zero kinetic energy with respect to you. But someone who is standing will see that it has kinetic energy! Now consider the following situation: a box of mass 12kg is pushed with a constant force so that so that its speed goes from zero to 1.5m/sec (as measured by the person at rest on the cart) and it covers a distance of 2.4m. Assume there is no friction. 2.4m v 1.5 /m s= mass of box 12 kg= 15 /m s docsity.com PHYSICS –PHY101 VU © Copyright Virtual University of Pakistan 20 h h / 2h A B C 0v D 2 2 2 2 Let's first calculate the change in kinetic energy: 1 (12 )(1.5 / ) 0 13.5 2 And then the (constant) acceleration: v v (1.5 / ) 0 2( ) f i f i f i K K K kg m s J m sa x x Δ = − = − = − − = = − 2 2 0.469 / 2(2 4 ) This acceleration results from a constant net force given by: (12 )(0.469 / ) 5.63 From this, the work done on the crate is: ( m s m F ma kg m s N W F x = ⋅ = = = = Δ = 5.63 )(2.4 ) 13.5 (same as 13.5 !)N m J K J= Δ = 5. Now suppose there is somebody standing on the ground, and that the trolley moves at 15 m/sec relative to the ground: 2 2 f i 2 2 Let us repeat the same calculation: 1 1 v v 2 2 1 1 (12 )(16.5 / ) (12 )(15.0 / ) 284 2 2 This example clearly shows that work and f iK K K m m kg m s kg m s J ′ ′ ′Δ = − = − = − = energy have different values in different frames. 6. The total mechanical energy is: . If there is no friction then is conserved. This means that the sum does not change with time. For example: a ball is thrown upwards at mech mechE KE PE E= + 0 2 2 0 0 speed v . How high will it go before it stops? The loss of potential 1 v energy is equal to the gain of potential energy. Hence, v . 2 2 Now look at the smooth, frictionless motion of m mgh h g = ⇒ = a car over the hills below: 15 /m s 15 /m s 16.5 /m s 50.4m 48.0m docsity.com
