Work and Kinetic Energy

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Whitney Fisher
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1 Work and Kinetic Energy Level : Physics I Teacher : Kim Objective Establish the relationship between work and energy Practice using Work-Kinetic Energy theorem compared to using ΣF=ma Understand how work&energy leads to the concept of simple machinery we use everyday Observe video clip on Work Introduction of Work done by Applied Force - Work(W) is defined as force applied in the direction of motion displacement W=Fcosφ d [J] Fapp φ d= distance traveled - Both force and distance is directly related to work. - Unit of work is expressed in Joules(J) - φ is the angle the between the direction of force and the direction of motion - If you apply 1N of force over a distance of 1meter, then you are doing 1Joule of work. Q1) A boy exerts a horizontal force of 10N on a box at a distance of 25m on a flat surface. How much work did he do on the box? a) 150J b) 200J c) 250J d) 300J Q2) A girl exerts a force of 35N at an angle of 35 above the horizontal along the handle of her wagon as she pulls it 25m across a level sidewalk. How much work does she do on her wagon? a) 698J b) 717J c) 756J d) 782J

2 Lifting a barbell is doing work. Two things happen when lifting a barbell i) a force is exerted on the barbell, and ii) the barbell is moved by that force Here, work is done on the barbell, since there is force being applied on the barbell and the barbell moves a certain distance Q3) A barbell is on the floor is lifted by man. If the mass of the barbell is 60kg, and he lifts it at a constant speed, 2m from the ground, how much work did he do on the barbell? a) 295J b) 437J c) 745J d) 1176J Q4) If the man is just holding the barbell above his head, how much work is done on the barbell? a) He is doing no work on the barbell. Since W=F d and distance covered is zero, so W=0. b) Of course he is doing work, since holding the barbell above his head causes strain on his body Work done by other Forces - Any force that may exist on an object has the potential to do work. In the figure below, three forces are acting on the block; F N, F g and F app. But only F app contributes to doing work on the block. (surface is frictionless) FN Fapp φ Fg=mg d= distance traveled => Work done by normal force F N is zero; W=F N cos90º d =0 => Work done by gravitational force F g(=weight) is zero; W=F g cos90º d = 0 *~ The only time a force does work on an object is when the force plays a role in moving the object

3 Work done by Gravity i) Vertical Motion Q5) When an object is free-falling in a vacuum, what force is doing work on that object? Fg ii) Incline Plane Q6) When an object is sliding down on the frictionless ramp, what force is doing work on that object? FN d Fg sinθ Fg θ *~Work-Kinetic Energy Theorem If an object is in motion, then it must have kinetic energy(k) K= 1 2 mv2 If you apply a force over a certain distance, there will be a change in kinetic energy vi distance d vf Fapp Fapp Ki + Fcosφφ d = Kf 1 2 mvi 2 +Fcosφ d = 1 2 mvf 2 Ki + ΣΣW = Kf : Work-Kinetic Energy Theorem - Unit of energy or work is Joules[J]

4 *~ Are we interested in the acceleration or the speed??~* W=Fcosφ d K= 1 2 mv2 K i + ΣΣW = K f A. On a Flat Surface Q7) If a 5kg object is sliding on a frictionless surface with a constant speed of 10m/s, i) What is the kinetic energy of the object? ii) If you apply a horizontal force of 5N over a distance of 10m, what is the new kinetic energy? Use ΣF=ma and Work-Kinetic Energy theorem and compare the process and result Newton s 2 nd Law : ΣF=ma *~ You need to also use v 2 f -v 2 i =2ad Work-Kinetic Energy theorem Q8) A 4kg object is sliding on a frictionless surface with a constant speed of 8m/s, and you apply a horizontal force of 8N over a distance of 12m. i) How much work was done by the applied force? a) 96J b) 82J c) 74J d) 68J ii) What is the new kinetic energy? a) 162J b) 178J c) 195J d) 224J iii) What is the new speed? a) 10.6m/s b) 8.4m/s c) 7.1m/s d) 5.9m/s

5 W=Fcosφ d K= 1 2 mv2 K i + ΣΣW = K f Q9) A 2.5kg object is sliding on a frictionless surface with a constant speed of 11.3m/s. If you apply a horizontal force of F p=6n, what is the speed of the object when it covered a distance of 21m? a) 14.75m/s b) 15.1m/s c) 19.4m/s d) 25.1m/s Q10) A 2.5kg object is sliding on a frictionless surface with a constant speed of 11.3m/s. If you apply a force of F p=6n at an angle of 27 above the surface, what is the speed of the object when it covered a distance of 21m? a) 14.75m/s b) 15.1m/s c) 19.4m/s d) 25.1m/s 27 Q11) An object of mass 2kg is sliding on a frictionless surface with a constant speed of 8m/s. How much force F p is needed to stop the object over a distance of 10m? (*~remember that the force is applied in the opposite direction of the motion for the object to stop~*) a) 3.9N b) 6.4N c) 8.3N d) 11.8N FN direction of motion Fg

6 W=Fcosφ d f=μf N K= 1 2 mv2 K i + ΣΣW = K f Q12) A 2.5kg box is resting on the flat rough surface. The coefficient of kinetic friction between the box and the floor is µ k =0.52.The box is then pulled by a force of F p=18n. fk FN direction of motion Fg i) How much work was done on the box by the pulling force F p through a distance of 5.2m? That is, find W p. a) 62.4J b) -62.4J c) 93.6J d) -93.6J ii) How much work was done on the box by the frictional force f k through a distance of 5.2m? That is, find W f. a) 66.25J b) J c) 25.48J d) J iii) What is the net work ΣΣW done on the box? That is, W p + W f. a) 27.4J b) 33.5J c) 49.7J d) 75.14J iv) What is the speed of the box when it covered 5.2m? a) 39.8m/s b) 27.2m/s c) 15.6m/s d) 4.7m/s Q13) A 4kg box is sliding on a rough surface with an initial speed of 8m/s. The coefficient of kinetic friction between the box and the floor is µ k =0.4. What is the speed of the box after it moved a distanced of 4m? (*~frictional force f=µf N, where F N=mg in this case ~*) a) 19.6m/s b) 14.2m/s c) 10.5m/s d) 5.7m/s fk FN direction of motion Fg

7 W=Fcosφ d f=μf N K= 1 2 mv2 K i + ΣΣW = K f B. Vertical Motion Use Work-Kinetic Energy to solve Q3) again! A barbell is on the floor is lifted by man. If the mass of the barbell is 60kg, and he lifts it at a constant speed, 2m from the ground, how much work did he do on the barbell? a) 295J b) 437J c) 745J d) 1176J C. Incline Plane Q14) A 8kg box is released from rest on a 30 inclined frictionless plane and accelerates down the incline. The distance from the box to the end of the ramp is 4m i) Find the work done by gravity through a distance of 4m down the ramp. a) 271.6J b) 156.8J c) 101.2J d) 59.7J ii) What is the speed of the box when it reaches the bottom of the ramp? a) 6.3m/s b) 4.37m/s c) 2.53m/s d) 1.9m/s Q15) A 13.2kg box is released from rest on an 11 inclined frictionless plane and accelerates down the incline. If the distance from the box to the end of the ramp is 5.1m, what is the speed of the box when it reaches the bottom of the ramp? a) 6.3m/s b) 4.37m/s c) 2.53m/s d) 1.9m/s

8 W=Fcosφ d f=μf N K= 1 2 mv2 K i + ΣΣW = K f Q16) A 6.3kg box is being pulled with a force of 56N from the bottom of the ramp starting from rest. The surface of the ramp is frictionless and the angle of the ramp is 40. i) How much work was done on the box by the pulling force F p through a distance of 5.2m? That is, find W p. a) 291.2J b) 233.9J c) 187.2J d) 105.6J ii) How much work was done on the box by gravitational force F g through a distance of 5.2m? That is, find W g. a) b) J c) 206.4J d) J iii) Find the speed of the box when it reaches a distance of 5.2m from the bottom. a) 26.9m/s b) 13.4m/s c) 5.19m/s d) 1.21m/s Q17) A 5.6kg box is being pulled with a force of 40N from the bottom of the ramp starting from rest. The surface of the ramp is frictionless and the angle of the ramp is 35. Find the speed of the box when it reaches a distance of 6.4m from the bottom. a) 2.56m/s b) 4.41m/s c) 6.78m/s d) 8.67m/s

9 W=Fcosφ d f=μf N K= 1 2 mv2 K i + ΣΣW = K f Q18) A 8kg box is released from rest on a 30 inclined rough plane and accelerates down the incline. The coefficient of friction µ k =0.54. i) If the distance from the block to the end of the ramp is 4m, what is the speed of the block when it reaches the bottom of the ramp? a) 3.3m/s b) 2.4m/s c) 1.6m/s c) 0.4m/s ii) The block will eventually slide off the ramp and continue on the surface. If the coefficient of kinetic friction on the flat surface was µ k =0.12, how far will the block travel until it stops? a) 2.21m b) 1.63m c) 1.07m d) 0.64m

10 W=Fcosφ d f=μf N K= 1 2 mv2 K i + ΣΣW = K f Q19) A 3.6kg box is being pulled with a force of 42N from the bottom of the ramp starting from rest. The surface is rough, so the coefficient of friction μ k=0.13, and the angle of the ramp is 26. i) Find the work done by the pulling force, W p, through a distance of 4.8m. a) 201.6J b) 148.4J c) 96.4J d) 62.2J ii) Find the work done by the gravitational force W g, through a distance of 4.8m a) 37.1J b) -37.1J c) 74.2J d) -74.2J iii) Find the work done by the frictional force, W f, through a distance of 4.8m a) 19.8J b) -19.8J c) 39.6J d) -39.6J iii) What is the net work ΣΣW done on the box? That is, W p + W g + W f. a) 46.5J b) 107.6J c) 168.4J d) 197.5J iv) Find the speed of the box when it reaches a distance of 4.8m from the bottom. a) 1.23m/s b) 3.18m/s c) 5.54m/s d)7.73m/s

Work and Kinetic Energy I

Work and Kinetic Energy I Scalar Product The scalar product of any two vectors A and B is a scalar quantity equal to the product of the magnitudes of the two vectors and the cosine of the angle φ between