Energy in two dimensions Subtitle

8m 5kg We’ve learned that we can measure the potential energy of an object by its distance from a reference level using gravity. Using W = Fd where gravity is our force we get gravitational potential energy: GPE = 8 * 10 * 5 = 400J Reference level

4m 2kg 30 0 In situations with 2 dimensions it is the same, though we will need our trigonometry rules to solve them. The reference level may not always be the ground. What would be the potential energy due to gravity in this situation? Reference level GPE = 2 * 10 * cos(30) * 4 = 69.28J

We can also deduce that PE due to gravity can be measured without regard to the slope. If the height is 100m and the block has a mass of 65kg, then GPE is still found as, no matter how long the slope is or even if it’s not straight: GPE = 65 * 10 * 100 = 65000J

Other problems may include a resisting force. For example, if this ramp is 150m long, the angle 10 0, the object 65kg, and the resisting force is 40N, then what is the total work done to move the box assuming it starts and stops at rest? Calculate work due to gravity: W = mgh = 65 * 10 * sin(10) * 150 W = 16931J Calculate work due to resistance: W = FD = 150 * 40N W = 6000 Total is 16931J J = 22931J

In other circumstances the velocity is not constant. Say for example Mr. Bach, who together with his bike has a mass of 90kg, starts riding up a hill at 50ms -1 ( 好厉害 ) but after 100m has slowed to 10ms -1. The hill is at a 15 0 incline. 1) What is his loss in KE? Ke initial =0.5 * 90 * 50 2 Ke initial = J Ke final = 4500J Ke loss = J

In other circumstances the velocity is not constant. Say for example Mr. Bach, who together with his bike has a mass of 90kg, starts riding up a hill at 50ms -1 ( 好厉害 ) but after 100m has slowed to 10ms -1. The hill is at a 15 0 incline. 2) What is his gain in PE? PE gain = 90 * 10 * sin(15) * 100 PE gain = 23294J

In other circumstances the velocity is not constant. Say for example Mr. Bach, who together with his bike has a mass of 90kg, starts riding up a hill at 50ms -1 ( 好厉害 ) but after 100m has slowed to 10ms -1. The hill is at a 15 0 incline. 3) If there is wind resistance with a force of 5N then how much work did Mr. Bach do? Work = PE gain + W resist – Ke loss W = 23294J + (5 * 100) – W = J

Example Exam Problem A lorry of mass kg climbs a straight hill ABCD which makes an angle ϴ  with the horizontal, where sin( ϴ )  = For the motion from A to B, the work done by the driving force of the lorry is 1200 kJ and the resistance to motion is constant and equal to 1240 N. The speed of the lorry is 15 ms −1 at A and 12 ms −1 at B. 1) What is the distance from A to B?

Example Exam Problem A lorry of mass kg climbs a straight hill ABCD which makes an angle ϴ  with the horizontal, where sin( ϴ )  = For the motion from A to B, the work done by the driving force of the lorry is 1200 kJ and the resistance to motion is constant and equal to 1240 N. The speed of the lorry is 15 ms −1 at A and 12 ms −1 at B. 2) What is the distance from B to D given that the lorry has a gain in potential energy of 2400kJ?

Example Exam Problem A lorry of mass kg climbs a straight hill ABCD which makes an angle ϴ  with the horizontal, where sin( ϴ )  = For the motion from A to B, the work done by the driving force of the lorry is 1200 kJ and the resistance to motion is constant and equal to 1240 N. The speed of the lorry is 15 ms −1 at A and 12 ms −1 at B. 3) For the motion from B to D the driving force of the lorry is constant and equal to 7200 N. From B to C the resistance to motion is constant and equal to 1240 N and from C to D the resistance to motion is constant and equal to 1860 N. Given that the speed of the lorry at D is 7 m s−1, find the distance BC.