Work and Energy. Work and Energy

Transcription

1 1. Work as Energy Transfer Work done by a constant force (scalar product) Work done by a varying force (scalar product & integrals). Kinetic Energy Work-Energy Theorem Work by a Baseball Pitcher A baseball pitcher is doing work on (or energy transfer to) the ball as he exerts the force over a displacement. = 0 = 44 m/s Key phrase: work done Work and by Energy Force on Body 1

2 Work (W) Work Done by a Constant Force (I) How effective is the force in moving a body? Both magnitude (F) and directions (θ ) must be taken into account. W [Joule] = ( F cos θ ) d = F d Review: Scalar Product (I) y AB = A B cosθ B = B cosθ AB = (A x i + A y j) (B x i + B y j) = (A x i + A y j) (B x i) +(A x i + A y j) (B y j) = (A x B x )(ii)+(a y B x ) (ji) +(A x B y )(ij)+(a y B y ) (jj) = (A x B x )(ii)+(a y B y ) (jj) x = A x B x + A y B y

3 Review: Scalar Product (II) y A B cosθ = A x B x + A y B y IF (A x, A y ) and (B x, B y ) are given, THEN you can determine the angle θ (between two vectors). x Work Done by a Constant Force (II) Example: Work done on the bag by the person.. Special case: W = 0 J a) W P = F P d cos ( 90 o ) b) W g = m g d cos ( 90 o ) Nothing to do with the motion 3

4 Example 1A A 50.0-kg crate is pulled 40.0 m by a constant force exerted (F P = 100 N and θ = 37.0 o ) by a person. A friction force F f = 50.0 N is exerted to the crate. Determine the work done by each force acting on the crate. Example 1B A 50.0-kg crate is pulled 40.0 m by a constant force exerted (F P = 100 N and θ = 37.0 o ) by a person. Assume a coefficient of friction force µ k = Determine the work done by each force acting on the crate and its net work. 4

5 Work-Energy Theorem W net = F net d = ( m a ) d = m [ ( 1 ) / d ] d = (1/) m (1/) m = K K 1 Work (W) Work as Energy Transfer [θ = constant, F = constant] Work Done by a Constant Force (I) How effective is the force in moving a body? Both magnitude (F) and directions (θ ) must be taken into account. Fcosθ Work W [Joule] = ( F cos θ ) d = F d W = K = K b - K a = ½ m ½ m Work-Energy Theorem a Distance, d b 5

6 Problem Solving Steps [Constant Force] (1) F.B.D. for the backpack () W i = F i d cosθ (3) W net = ΣW i (4) W net = K Analyze the motion Example 1C A hiker carries a 15.0-kg backpack up a hill of height h = 10.0 m. Determine (a) the work done by the hiker on the backpack and (b) the work done by the gravity. For simplicity, assume the motion is smooth and at constant velocity. (1) F.B.D. for the backpack () W i = F i d cosθ (3) W net = ΣW i (4) W net = K Analyze the motion Key phrase: work done by Work A and on Energy B 6

7 Example A car traveling 60.0 km/h to can brake to a stop within a distance of 0.0 m. If the car is going twice as fast, 10 km/h, what is its stopping distance? (a) (b) (Part II) 1. Work Energy as Energy Transfer Work done by a constant force (scalar product) Work done by a varying force (scalar product & integrals). Kinetic Energy Work-Energy Theorem 7

8 Work as Energy Transfer [Theomodynamics] (1) F.B.D. () W by each force (3) W net Example 3 motion 1 = 0 =? d = 5.00 m µ k = (4) W-E Theorem to find. 8

9 Work (W) Work as Energy Transfer [Area in F-d Graph] Work Done by a Constant Force (I) How effective is the force in moving a body? Both magnitude (F) and directions (θ ) must be taken into account. Fcosθ Work W [Joule] = ( F cos θ ) d = F d W = K = K b - K a = ½ m ½ m Work-Energy Theorem a Distance, d b Work (W) Work: Area in F-d Graph [F = varied, θ = constant] Work Done by a Constant Force (I) How effective is the force in moving a body? Both magnitude (F) and directions (θ ) must be taken into account. W [Joule] = ( F cos θ ) d = F d Fcosθ W = K = K b - K a = ½ m ½ m Work-Energy Theorem Work Distance, d 9

10 Work Done by a Varying Force (direction and magnitude) W 1 = F 1 cosθ 1 l 1 F = F 1 cosθ 1 Fcosθ Distance, l Motion a b is collection of straightline motion with a constant force Problem Solving Steps [Varying Force] (1) F.B.D. for the backpack () W i = F dl (3) W net = ΣW i (4) W net = K Analyze the motion 10

11 Spring Force (Hooke( Hooke s Law) [θ = constant, F = varied] Spring Force (Restoring Force): The spring exerts its force in the direction opposite the displacement. F S (x) = k x F S F P Natural Length x > 0 x < 0 0 x Work Done to Stretch a Spring F P (x) = k x F S F P Natural Length Final position W P = x W P = F P (x) d x = ½ k x ½ k x 1 x 1 x 1 F P (x) dx W Initial position 11

12 Example 1 A person pulls on the spring, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the person do? If, instead, the person compresses the spring 3.0 cm, how much work does the person do? Example A person pulls on the spring, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the spring do? If, instead, the person compresses the spring 3.0 cm, how much work does the spring do? 1

13 Example 3 A 1.50-kg block is pushed against a spring (k = 50 N/m), compressing it 0.00 m, and released. What will be the speed of the block when it separates from the spring at x = 0? Assume µ k = (i) F.B.D. first! (ii) x < 0 F S = k x Example 4 A 1.50-kg block is pushed against a spring, compressing it 0.00 m, and released. The spring force is given as F S (x) = kx + bx where k = 50 N/m and b = 300 N/m. What will be the speed of the block when it separates from the spring at x = 0? Assume µ k = (i) F.B.D. first! (ii) x < 0 F S = k x+bx 13