Work and Energy Problems A force F of strength 20N acts on an object of mass 3kg as it moves a distance of 4m. If F is perpendicular to the 4m displacement,

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Presentation transcript:

Work and Energy Problems

A force F of strength 20N acts on an object of mass 3kg as it moves a distance of 4m. If F is perpendicular to the 4m displacement, the work done is equal to: a)0J b)60J c)80J d)600J e)2400J Multiple Choice Since the Force is perpendicular to the displacement, there is no work done by this force.

Multiple Choice This is a job for the work energy theorem. Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to 6 m/s in 8s. How much work was done on the object during this time? a)27J b)54J c)72J d)96J e)Can not be determined from the information given

Multiple Choice The length of the ramp is irrelevant, it is only the change in height A box of mass m slides down a frictionless inclined plane of length L and vertical height h. What is the change in gravitational potential energy? a)-mgL b)-mgh c)-mgL/h d)-mgh/L e)-mghL

Multiple Choice Since centripetal force always points along a radius towards the centre of the circle, and the displacement is always along the path of the circle, no work is done. An object of mass m is travelling at constant speed v in a circular path of radius r. how much work is done by the centripetal force during one-half of a revolution? a)πmv 2 J b)2πmv 2 J c)0J d)πmv 2 r J e)2πmv 2 /r J

Multiple Choice The gravitational force points downward, while the displacement if upward While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above the floor, how much work does the gravitational force do on the block? a)-30J b)-15J c)0J d)15J e)30J

Multiple Choice The work done by gravity is equal to the change in Kinetic Energy. A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that makes an angle of 30 o with the horizontal. If the block is released from rest at the top of the incline, what is the speed at the bottom? a)4.9 m/s b)5.2 m/s c)6.4 m/s d)7.7 m/s e)9.2 m/s

Multiple Choice Use the power equation A force of 200N is required to keep an object at a constant speed of 2 m/s across a rough floor. How much power is being expended to maintain this motion? a)50W b)100W c)200W d)400W e)Cannot be determined with given information

Question A truck moves with velocity v 0 = 10 m/s on a slick road when the driver applies the brakes. The wheels slide and it takes the car 6 seconds to stop with a constant deceleration. a) How far does the truck travel before stopping? b) Determine the kinetic friction between the truck and the road.

Solution to Question A truck moves with velocity v 0 = 10 m/s on a slick road when the driver applies the brakes. The wheels slide and it takes the car 6 seconds to stop with a constant deceleration. a) How far does the truck travel before stopping? b) Determine the kinetic friction between the truck and the road.

Solution to Question A truck moves with velocity v 0 = 10 m/s on a slick road when the driver applies the brakes. The wheels slide and it takes the car 6 seconds to stop with a constant deceleration. a) How far does the truck travel before stopping? b) Determine the kinetic friction between the truck and the road. First we need the acceleration the car experiences. Since the only acceleration force experienced by the car is friction

Example An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the coefficient of kinetic friction is Use the Work Energy Theorem to determine the distance the object travels before coming to a halt. 1.0 kg

Solution An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the coefficient of kinetic friction is Use the Work Energy Theorem to determine the distance the object travels before coming to a halt. Forces mg FNFN FfFf We can see that the objects weight is balanced by the normal force exerted by the ice. Therefore the only work done is due to the friction acting on the object. Let’s determine the friction force. Now apply the work Energy Theorem and solve for d

Bouncing Ball In a perfect world a 1.02 kg ball is dropped from 2.0 m and all the energy is converted from potential to kinetic so each time the ball bounces it reaches its original height. 1) Find the velocity of the ball after the first drop as the ball reaches the ground? E M = Ep + Ek v = 0 m/s at the top. E M Top = mgh + ½ mv 2 E M Bottom = Ep + ½ mv 2 20 J = ½ (1.0kg) v 2 E M = Ep + Ek E M Half = mgh + ½ mv 2 = 10 J + 10 J = 20 J E M = Ep + Ek E M Half = mgh + ½ mv 2 = 10 J + 10 J = 20 J = 1.02 kg (9.8 m/s 2 ) (2m) = 20 J Hey What would the kinetic energy be half- way down?

What speed will a 200g baseball reach when dropped from a height of 2.0 m? Would the speed change if the ball was bowling ball (5.0kg)? Given: m = 200 g = 0.200kg h 1 = 2.0 m h 2 = 0 m g = 9.8 m/s 2 v 1 = 0 m/s V 1 = 0 m/s h 2 = 0 m

Energy Efficiency

Ex. A firefly’s has 4.13 J of chemical food energy that is converted to 3.63 J of glowing radiant energy. What is the fly’s efficiency of its energy transformations?

Ex. What is the efficiency of a rope and pulley system if a painter used 1.93 kJ of mechanical energy to pull on the rope and lift a 20.0 kg paint barrel at a constant speed to a height of 7.5 m? Given: E = kJ = 1930 J h 1 = 0 m h 2 = 7.5 m m = 20.0 kg g = 9.8 m/s 2 ΔEp = W net

Collective Energy: Power in Numbers??? (for fun) Vsauce:

A Powerful Question How much power does a swimmer produce if she transforms 2.4 kJ of chemical energy into kinetic and thermal energy in 12.5 s? P = ΔE / Δt = 2400 J / 12.5s = 192 W Δ E = 2400 J Δ t = 12.5 s P = ?

An elevator and its occupants have a mass of kg. The elevator motor lifts the elevator to the 12 th floor, a distance of 40.0m in 74.8s. What is the average power output of the motor? Given: m = 1300 kg h = 40.0 m g = 9.81 m/s 2 Δt = 74.8 s P = ΔE / Δt = mgh / Δt = (1300.0kg)(9.81m/s2)(40.0m) 74.8 s = 6810 J/s = 6810 W

What is the cost of operating a 60.0 W light bulb 4.0 hours a day for 6.0 days if the price of electrical energy is $0.13/kWh? 1 kWh = 1000 W x h ΔE = P x Δt h = 4.0 h/day x 6.0 days = 24h Cost = 1.44 kWh x 0.13/kWh = $ = $0.19 E = 60.0 W x 24 h = 1440 Wh = 1.44 kWh

Example A box of mass 5kg is released from rest at point A, the top of a long frictionless slide. Point A is at height 10m above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The friction Force between the box and this surface is 2N, and the horizontal distance between points B and C is 3m. BC x U k. H A a)Find the speed of the box when its height above the ground is 1/2H b)Find the speed of the box when it reaches B. c)Determine the work done by friction between B and C. d)Determine the speed of the box when it reaches C e)If the slide was not frictionless, determine the work done by friction as the box moved from A to B if the speed at B was ½ of the speed calculated in b)

Example A box of mass 5kg is released from rest at point A, the top of a long frictionless slide. Point A is at height 10m above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The friction between the box and this surface is 2N, and the horizontal distance between points B and C is 3m. a) Find the speed of the box when its height above the ground is 1/2H BC x ukuk H A

Example A box of mass 5kg is released from rest at point A, the top of a long frictionless slide. Point A is at height 10m above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The friction between the box and this surface is 2N, and the horizontal distance between points B and C is 3m. b) Find the speed of the box when it reaches B. BC x ukuk H A

Example A box of mass 5kg is released from rest at point A, the top of a long frictionless slide. Point A is at height 10m above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The friction Force between the box and this surface is 2N, and the horizontal distance between points B and C is 3m. c) Determine the work done by friction between B and C BC x ukuk H A Hey, note that the work done by friction is negative, it took away 6J of energy.

Example A box of mass 5kg is released from rest at point A, the top of a long frictionless slide. Point A is at height 10m above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The friction Force between the box and this surface is 2N, and the horizontal distance between points B and C is 3m. d) Determine the speed of the box when it reaches C. BC x ukuk H A L Why, use mgh for K B ?

Example A box of mass 5kg is released from rest at point A, the top of a long frictionless slide. Point A is at height 10m above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The friction Force between the box and this surface is 2N, and the horizontal distance between points B and C is 3m. e) If the slide was not frictionless, determine the work done by friction as the box moved from A to B if the speed at B was ½ of the speed calculated in b) B x ukuk H A