Potential Energy & Conservation of Energy Physics

Transcription

1 Potential Energy & Conservation of Energy Physics Work and Change in Energy If we rearrange the Work-Kinetic Energy theorem as follows Ki +Fcosφ d = Kf => Fcosφ d = Kf - Ki => Fcosφ d = K => Ki + ΣΣW = Kf => ΣΣW = K If you apply a force over a certain distance on an object or in other words, if we do work on an object, there will be a change in kinetic energy In general, if work is done on an object, there will be a change in energy of some type Work = Energy - If doing work results in a change in energy (usually an increase in energy), then if an object has energy, it means that object has the ability to do work => We need energy to do work = To do work, we need energy *~ Observe Video on Kinetic Energy & Potential Energy Gravitational Potential Energy - When an object is lifted and placed on top of a wall, we are doing work on that object. (lifting force vertical distance = work) - If the object started from rest and placed on top of the wall, then Ki = Kf, so there is no change in kinetic energy. F lift - However, if work is done on the object, there is always bound to be change in energy on that object of some type. - The energy we are changing is called Gravitational Potential Energy. The symbol for gravitational potential energy is Ug. 1

2 Lifting an object means doing work on the object - Exerting a force over a certain distance - More distance the block is lifted, the more work is being done - The more work is being done on the object means having more energy - The higher the object is being lifted, the more energy is being changed (in this case, increased) => Gravitational potential energy is directly proportional to the vertical distance(=height) Gravitational Potential energy (Ug ) = mgh where m is the mass, g=9.8m/s, h is height and the unit for energy is Joules[J] - Gravitational potential energy is often called energy of position (vs kinetic energy ~ v) Potential Energy in General - Whenever you do work against a restoring force, you are storing energy. - Lifting something means applying a force against the force of gravity that is trying to pull the object back down to its original position - This stored energy can be used for later use. Hence, it is called potential energy. - If an object has potential energy, once released, that stored energy will be converted into kinetic energy **Conservation of Mechanical Energy Energy cannot be created or destroyed. It can be transformed from one form into another, but the total amount of energy never changes Q1) If you drop a ball from a certain height, does Ug increases or decrease? a) Increase b) decrease c) stays the same Q) When a ball is falling, does the kinetic energy increase or decrease? a) Increase b) decrease c) stays the same The sum of two energies, kinetic and gravitational potential energy is called total Mechanical Energy (E) When an object falls, the potential energy decreases while the kinetic energy increases, but the total mechanical energy remains constant throughout the motion, if there was no external force acting on the system E = Ki + Ugi = Kf + Ugf

3 If you drop a ball from a certain height, the gravitational potential energy(ug) decreases and the kinetic energy(k) increases, but the total mechanical energy is conserved. *~ U g = mgh K = 11 mmvv ~* Ug =1000J K=0J Total Energy = Ug + K = 1000J h Ug =500J K=500J Total Energy = Ug + K = 1000J Ug =50J K=750J Total Energy = Ug + K = 1000J Ug =0J K=1000J Total Energy = Ug + K = 1000J K = 1 mv² Ug = mgh Ki + Ugi = Kf + Ugf Q3) An object of mass 10kg is lifted to a height of 10m. i) What is the gravitational potential energy of the object at that height? What is the kinetic energy at that height? ii) When the object falls 5m, what is the gravitational potential energy and kinetic energy at that height? What is the speed at that height? iii) When the object falls 8m, what is the gravitational potential energy and kinetic energy at that height? What is the speed at that height? 3

4 K = 1 mv² Ug = mgh Ki + Ugi = Kf + Ugf Practice Questions Q4) An apple with a mass of 0.1kg falls off a tree branch with a height of.75m. Find the velocity of the apple when it hits the ground? a) 7.3m/s b) 8.8m/s c) 9.4m/s d) 1.4m/s Q5) A bowling ball with a mass of 5kg falls off a shelf with a height of m. How much kinetic energy would the bowling ball have when it has fallen halfway to the floor? a) 8J b) 69J c) 53J d) 49J Q6) A 0.04kg body starting from rest falls through a vertical distance of 0.5m to the ground. (i) What is the kinetic energy of the body just before it hits the ground? (ii) What is the velocity of the body just before it hits the ground? a) 0.098J, 4.45m/s b) 0.098J,.1m/s c) 0.049J, 4.45m/s d) 0.049J,.1m/s Q7) A boy throws a 0.15kg stone upward from the top of a 0m cliff with a speed of 15m/s. Find its kinetic energy and speed when it lands in a river below. a) 1.4J, 16.3m/s b) 5.J, 18.4m/s c) 46.3J, 4.8m/s d) 53.5J, 9.5m/s Ug,max, K=0 Ug + K Ug=0, Kmax 4

5 K = 1 mv² Ug = mgh Ki +ΣW=Kf Ki + Ugi = Kf + Ugf Energy is path independent Q8) A block is released from the top of the frictionless ramp. The height of the ramp is h=5m for all three cases. Find the speed of the block at the bottom of the ramp for each case. d is the distance of the ramp. Use Ki +ΣW=Kf and Ki + Ugi = Kf + Ugf d=10m d=7.08m d=5.78m θ=30 θ=45 θ=60 case-1 case- case-3 Q9) A child rides on a smooth slide of a height of m. The child starts from rest at the top. Determine the speed at the bottom. 5

6 K = 1 mv² Ug = mgh Ki +ΣW=Kf Ki + Ugi = Kf + Ugf Q10) A roller coaster starts from rest at the top of an 18m hill. The car travels to the bottom of the hill and continues up the next h=18m hill that is 10m high. How fast is the car moving at the top of the hill if the friction is ignored? a) 6.4m/s b) 8.1m/s c) 1.5m/s d) 18.7m/s h=10m Q11) A roller coaster starts from rest at position 1 shown below and rolls without friction along the loop. The roller coaster first pass A, goes around the loop and passes B and then passes C. The height h=1m and the radius of the loop is r=0.4m. Find the speed of the roller coaster at position A, B and C. h B r A C Ramp Revisit Q1) You want to lift an object of mass 50kg to height of 5m. By using a frictionless ramp(incline angle is θ=11.5 ), you can decrease the force by applying the force at a longer distance. i) How much work was needed to lift the object straight up at constant speed?(=wl). ii) How much work was needed to push the object along the ramp to the same height at constant speed?(=wp) iii) How much force was needed to push the object along the ramp?(=fp) F p F L 6