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Work Done by a Constant Force

Work Done by a Constant Force. Work W  F || d = Fd cos θ For a CONSTANT force!. Consider an object moving in straight line. It starts at speed v 1 . Due to the presence of a net force F net , ( ≡ ∑ F) , it accelerates (uniformly) to speed v 2 , over a distance d .

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Work Done by a Constant Force

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  1. Work Done by a Constant Force Work W  F||d = Fd cosθ For a CONSTANT force!

  2. Consider an object moving in straight line. It starts at speed v1. Due to the presence of a net force Fnet, (≡∑F), it accelerates (uniformly) to speed v2, over a distance d. Newton’s 2nd Law: Fnet= ma (1) (v2)2 = (v1)2 + 2ad (2) What is the relationship between Work and Energy? Hint: Combine the 2 formulas to solve for Work (W)

  3. Wnet = (½)m(vf)2 – (½)m(vi)2 Work-Energy Formula Kinetic energy associated with the motion of an object Scalar quantity with the same unit as work Derive the Units!!!

  4. 1) A truck moving at 15 m/s has a kinetic Energy of 1.4 x 105 J. What is the mass of the truck? 2) How much work is required for a 74 kg sprinter to accelerate from rest to 2.2 m/s? Practice Problems

  5. Kinetic energy & work done on a baseball A baseball, mass m = 145 g (0.145 kg) is thrown so that it acquires a speed v = 25 m/s. a. What is its kinetic energy? b. What was the net work done on the ball to make it reach this speed, starting from rest?

  6. Work on a car to increase its kinetic energy Calculate the net work required to accelerate a car, mass m = 1000-kg car from v1 = 20 m/s to v2 = 30 m/s.

  7. Conceptual Example:Work to stop a car A car traveling at speedv1 = 60 km/hcan brake to a stop within a distanced = 20 m. If the car is going twice as fast,120 km/h, what is its stopping distance? Assume that the maximum braking force is approximately independent of speed.

  8. Wnet = Fd cos (180º) = -Fd(from the definition of work) Wnet = KE = (½)m(v2)2 – (½)m(v1)2(Work-Energy Principle) but, (v2)2 = 0(the car has stopped) so-Fd = KE= 0 - (½)m(v1)2 or d  (v1)2 So the stopping distance is proportional to the square of the initial speed! If the initial speed is doubled, the stopping distance quadruples! Note:KE  (½)mv2  0 Must be positive, since m & v2 are always positive (real v).

  9. Solving Work Problems • Sketch a free-body diagram. • Choosea coordinate system. • ApplyNewton’s Laws to determine any unknown forces. • Find the work done by a specific force. • Findthe net work by either a. Find the net force & then find the work it does, or b. Find the work done by each force & add.

  10. Work Done by Multiple Forces If more than one force acts on an object, then the total work is equal to the algebraic sum of the work done by the individual forces Remember work is a scalar, so this is the algebraic sum

  11. Example The force shown has magnitude FP = 20 N & makes an angle θ = 30° to the ground. Calculate the work done by this force when the wagon is dragged a displacement d = 100 m along the ground.

  12. Work and Multiple Forces Suppose µk = 0.200, How much work done on the sled by friction, and the net work if θ = 30° and he pulls the sled 5.0 m ?

  13. Suppose µk = 0.200, How much work done on the sled by friction, and the net work if θ = 30° and he pulls the sled 5.0 m ?

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