Survey

Survey

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Work 1. Work don by a constant force Definition: Units: W F|| d Fd cos Fd [W] = N*m = J 2. Positive and negative work Work done by forces that oppose the direction of motion will be negative. Centripetal forces do no work, as they are always perpendicular to the direction of motion. Example: The object of mass m = 800 g is displaced by a constant the force F = 20 N as shown below (the angle θ=60º). The force of friction is f = 6 N. Find the work done by each of these forces and the total work. y Fy F Fx f a x 5m WF Fx x ( F cos ) x (20 N )(cos 600 )(5 m) 50 J W fr f (cos1800 ) x fx (6 N )(5 m) 30 J Wtot F cos f x 20 N cos 600 6 N (5m) 20 J Example: A block slides down a rough inclined surface. The forces acting on the block are depicted below. The work done by the frictional force is: A. Positive B. Negative fk C. Zero N fk 90° Δx Δx 180° mg W = |fk| |Δx| cos(180°) = -|fk| |Δx| < 0 Work done by the normal force: WN = |N| |Δx| cos(90°) = 0 0 < θ < 90° Work done by weight: Wmg = |mg| |Δx| cos(θ ) > 0 Work kinetic energy principle v22 v12 2a( x2 x1 ) mv22 mv12 ma( x2 x1 ) Fd 2 2 Definition: K mv 2 mv22 mv12 W 2 2 W=K2 - K1 2 Example: An 80-g arrow is fired from a bow whose string exerts an average force of 100 N on the arrow over a distance of 49 cm. What is the speed of the arrow as it leaves the bow? m = 80 g F = 100 N d = 49 cm v1= 0 v2 - ? W Fd K1 0 mv22 K2 2 mv22 Fd 2 2 Fd v2 m 2 100 N 49 102 m v2 35m / s 3 80 10 kg Example: Paul pushes a box along 10 m across the floor at a constant velocity by exerting a force of 200N. N Δx FPaul fk FPaul f k ma 0 FPaul f k mg •Work by Paul on the box: WPaul = (200 N)(10 m) = 2000 J (energy coming from the biochemical reactions in his muscles) •Work by friction on the box: Wf = (-200 N)(10 m) = -2000 J (released as thermal energy into the air and the floor) •WKE theorem: WPaul + Wf = 0 Kf – K i = 0 •The weight of the box that Paul pushes along a horizontal surface does no work since the weight is perpendicular to the motion. •The normal by the floor does no work either. Example: A ball is dropped and hits the ground 50 m below. If the initial speed is 0 and we ignore air resistance, what is the speed of the ball as it hits the ground? We can use kinematics or… the WKE theorem Work done by gravity: mgh W K mgh 12 mv 2 0 v 2 gh 2(9.8 m/s2 )(50 m) 31 m/s Example: Two blocks (m1=2m2) are pushed by identical forces, each starting at rest at the blue vertical line (start). Which object has the greater kinetic energy when it reaches the green vertical line (finish)? A. Box1 B. Box 2 C. They both have the same kinetic energy Same force, same distance Same work Same change in kinetic energy Example: What is the speed of the system after box 1 has fallen for 30cm? =x Two approaches: 1. Use Newton’s laws to find acceleration; use kinematics to find speed. (long!!) 2. Use WKE theorem (smart!) External forces doing work: m1g, fk (The tensions are internal forces: their net work is zero) 1 1 2 2 KE mv m v 0 1 2 2 2 Wnet m1 gx fk x m1 gx k m2 gx 1 m1 k m2 gx m1 m2 v 2 2 N m2 = 5.0 kg T fk m1 k m2 v 2 gx m1 m2 T m1 = 3.0 kg V=1.2 m/s m1g m2g μk = 0.2 Example: A person pulls a 10-kg block up a ramp of height h = 10 m and angle θ = 10 at constant speed. The rope makes an angle φ = 15 with the ramp and has a tension of 50 N. How many forces are doing work on the box? Find the work done by all the forces. A. 1 B. 2 Wg < 0 C. 3 Wf < 0 D. 4 E.5 WT > 0 WN = 0 WNet = 0 Δx (Constant speed) φ T Δx 1. Work by the tension: WT T x cos (50 N)(56.7 m) cos15 2740 N N φ T h fk θ x h / sin mg 2. Work by gravity: Δx Wg mg x cos 2 mg x sin mgh 980 J θ + π/2 mg Of course! –h is the component of the displacement in the direction of weight! 3. Work by the normal: WN =0 90 Δx N A person pulls a 10-kg block up a ramp of height h = 10 m and angle θ = 10 at constant speed. The rope makes an angle φ = 15 with the ramp and has a tension of 50 N. Find the work done by all the forces. 4. Work by the friction: Two options: 1. Find fk with Newton’s laws and use it to find the work (long) 2. Use the WKE theorem (smart): Wnet KE 0 Wnet Wg WN WT Wf 0 Wf Wg WN WT 980 0 2740 1760 J

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