Energy IN = Energy OUT Means ALL Energy 3.2.4 Work to Overcome Friction

We now consider problems in which Q ≠ 0 Energy “Loss” Energy can never by truly destroyed, but it can be converted into non-mechanical forms (mainly heat) that are not ‘useful’. We now consider problems in which Q ≠ 0

Example #1 – Lifting a Load 5200 joules of work are done in lifting a 50 kilogram object a distance of 10 meters. Determine the amount of potential energy gained by lifting this object. In the “real world” it takes more energy to lift an object than the gain in gravitational PE would suggest. ΔPE = mgΔh ΔPE = (50 kg)(9.81 m/s2)(10 m) ΔPE = 4905 J

Example #1 – Lifting a Load 5200 joules of work are done in lifting a 50 kilogram object a distance of 10 meters. How much energy must have been used to overcome friction? W = Fd = ΔET = 5200 J ET = PE + KE + Q 5200 J = 4905 J + 0 + Q Q = 295 J

Example #2 – Missing KE A 4.0 kilogram object begins at rest at the top of an incline that is 2.0 meters above the ground and slides down to the bottom. The object reaches a speed of 5.0 meters per second at the bottom of the incline. Determine the total energy of the object at the top of the incline. In the “real world” potential energy is not completely converted into kinetic energy when objects lose height. ET = PE + KE + Q ET = mgh + ½ mv2 + Q ET = (4.0 kg)(9.81 m/s2)(2.0 m) + 0 + 0 ET = 78.48 J

Example #2 – Missing KE Determine the amount of energy used to overcome friction as the block slides down the incline. ET = PE + KE + Q 78.48 J = mgh + ½ mv2 + Q 78.48 J = 0 + ½ (4 kg)(5.0 m/s)2 + Q Q = 28.48 J

Example #3 – Constant Velocity A 2.0 kilogram object is pulled along a flat surface at a constant speed with a force of 30 newtons. Calculate the amount of work done in moving the object 4.0 meters. In the “real world” we must constantly add energy to a system (do work) to keep things moving at a constant speed. W = Fd = ΔET W = (30 N)(4.0 m) W = 120 J

Example #3 – Constant Velocity A 2.0 kilogram object is pulled along a flat surface at a constant speed with a force of 30 newtons. How much work must have been done against friction in moving this object? W = Fd = ΔET = 120 J ET = PE + KE + Q 120 J = 0 + 0 + Q Q = 120 J

End of 3.2.4 - PRACTICE