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Work Done by Non-conservative Forces

Work Done by Non-conservative Forces. CH - 09. Work Done by Nonconservative Forces. Nonconservative forces change the amount of mechanical energy in a system. W nc = work done by nonconservative force.

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Work Done by Non-conservative Forces

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  1. Work Done by Non-conservative Forces CH - 09

  2. Work Done by Nonconservative Forces Nonconservative forces change the amount of mechanical energy in a system. Wnc = work done by nonconservative force

  3. Problem: Children and sled with mass of 50 kg slide down a hill with a height of 0.46 m.  If the sled starts from rest and has a speed of 2.6 m/s at the bottom, how much thermal energy is lost due to friction (i.e. what is the work that friction does)?  If the hill has an angle of 20° above the horizontal what was the frictional force.

  4. Since vi = 0, and hf = 0, The force done by friction is determined from;

  5. Example: A child of mass m rides on an irregularly curved slide of height h = 2.00 m, as shown in Figure. The child starts from rest at the top. (A) Determine his speed at the bottom, assuming no friction is present.

  6. (B) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf =3.00 m/s and m = 20.0 kg.

  7. Example: A block having a mass of 0.80kg is given an initial velocity vA = 1.2m/s to the right and collides with a spring of negligible mass and force constant k = 50N/m, as shown in Figure. (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision.

  8. (B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 0.50. If the speed of the block at the moment it collides with the spring is vA = 1.2 m/s, what is the maximum compression xC in the spring?

  9. Problem: A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s (Fig. P8.33). The block comes to rest after traveling 3.00m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine • the change in the block’s kinetic energy, • (b) the change in the potential energy of the block–Earth system, • (c) the friction force exerted on the block (assumed to be constant). • (d) What is the coefficient of kinetic friction?

  10. Problem: A 1300-kg car drives up a 17.0-m hill. During the drive, two nonconservative forces do work on the car: • the force of friction, and • the force generated by the car’s engine. • The work done by friction is –3.31  105 J; • the work done by the engine is +6.34  105 J. • Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill.

  11. Problem :A skateboard track has the form of a circular arc with a 4.00 m radius, extending to an angle of 90.0° relative to the vertical on either side of the lowest point, as shown in the Figure. A 57.0 kg skateboarder starts from rest at the top of the circular arc. What is the normal force exerted on the skateboarder at the bottom of the circular arc? What is wrong with this picture? At bottom, a = v2/r Which direction?

  12. Problem : A 1.9-kg block slides down a frictionless ramp, as shown in the Figure. The top of the ramp is 1.5 m above the ground; the bottom of the ramp is h = 0.25 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Find the distance d.

  13. Chapter (9) Review

  14. Example 1:A 65-kg athlete runs a distance of 600 m up a mountain inclined at 20o to the horizontal.  He performs this feat in 80s. Assuming that air resistance is negligible, (a) how much work does he perform and (b) what is his power output during the run?

  15. Solution • Assuming the athlete runs at constant speed, we have • WA + Wg = 0 • where WA is the work done by the athlete and Wg is the work done by gravity.  In this case, • Wg = -mgs(sinθ) • So • WA = -Wg  = + mgs(sinθ) •                              = (65kg)(9.80m/s2)(600m) sin20o • (b) His power output is given by

  16. Gravitational Potential Energy Gravitational potential energy U = mg(y-y0)y = height U=0 at y=y0 (e.g. surface of earth). Work done by gravity: Wg = mg Dy = mg (y- y0)

  17. Mechanical Energy (Conservative Forces) Mechanical energyE is the sum of the potential and kinetic energies of an object. E = U + K The total mechanical energy in any isolated system of objects remains constantif the objects interact only through conservative forces: E = constant Ef =Ei Uf + Kf = Ui+ Ki DU + DK = DE = 0

  18. Example: A ball of mass m is dropped from a height h above the ground, as shown in Figure (A) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground. (B) Neglecting air resistance, determine the speed of the ball when it is at the ground.

  19. Example:  A 0.5 kg block is used to compresses a spring with a spring constant of 80.0 N/m a distance of 2.0 cm.  After the spring is released, what is the final speed of the block? Solution

  20. Example: A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure. Determine (a) the particle’s speed at points B and C and (b) the net work done by the gravitational force in moving the particle from A to C.

  21. (a) the particle’s speed at points B Find the particle’s speed at points C then find the work from the relation

  22. For the car rolling up the incline

  23. Energy is conserved.To get a better idea about what potential energy is all about, consider the work energy theorem, Potential energy is only associated with conservative forces. In fact, this is the standard definition of potential energy.

  24. Example (1) :A 100g car is taken from the base to the top of a 10.0cm high ramp. FIND 1-Find the total work done by gravity as the car is taken up the ramp down the vertical side and back horizontally to its starting point.2- The speed at the bottom can be found using the Law of Conservation of Energy.

  25. 1-

  26. 3. Two Types of Potential EnergyEach conservative force has a potential energy associated with it. In particular, let’s look at gravityand springs.Gravitational Potential Energy: 1Applying the definition of potential energy to a ball and using the definition of work,

  27. Since the path doesn’t matter let’s go straight upward. In this case the weight and the displacement are opposite. Using the mass/weight rule,

  28. Example : 00 A pumpkin (نبات القرع) falls 62.0m to the ground. Find the speed just before it lands. • This problem could be solved using kinematics, but let’s use the Law of Conservation of Energy. Choose the coordinates • so that y=0 at the ground. There for, the potential energy is zero there.

  29. This is the same result as the kinematic equations. Plugging in the numbers,

  30. Spring Potential Energy:

  31. Example 7.5: A 5.00N weight is hung on a spring that drops a maximum of 1.02m. Find thespring constant.Choosing the coordinates so that y=0 corresponds tothe place where the spring is fully extended, the kinetic and potential energies can be found.

  32. Example: A 58-kg skier is coasting down a slope inclined at 25° above the horizontal. A kinetic frictional force of 70 N opposes her motion.Near the top of the slope the skier's speed is 3.6 m/s. What is her speed at a point which is 57 m downhill? Solution: the net-work = Wf+Wg :

  33. Now use the Work-Energy Theorem:

  34. Example: A 65-kg bicyclist rides his 10.0-kg bicycle with a speed of 12 m/s. • How much work must be done by the brakes to bring the bike and rider to a stop? • How far does the bicycle travel if it takes 4.0 s to come to rest? • What is the magnitude of the braking force? Solution: • Friction = only horizontal force • Wnet = Kf -Ki <0 • Wnet = 0 – (1/2) m vi2 • Wnet = - (1/2) (10+65) (12)2 • Wnet = - 5400 kg m 2 /s 2 = -5400 J

  35. Find acceleration first • v = v0+at • 0 = v0 + at, • a= -v0/t • a = -12/4 = -3 m/s2 • x= x0+v0 t+(1/2)at2 • x= 0 + (12)(4) + (1/2)(-3)(4)2 • x= 48 -24 = 24 m • Braking force • Wnet = Fnet d = FFriction d • FFriction = Wnet /d = (-5400)/(24) = -225 N

  36. 7.7: A 10.0kg block slides down a 3.00m high frictionless ramp. Then it skids 6.00malong a rough surface before returning to a smooth surface and colliding with a spring of spring constant 2250N/m. The block comes to rest when the spring is compressed 30.0cm. Find thecoefficient of friction between the block and the rough surface.

  37. Example: A child of mass m rides on an irregularly curved slide of height h = 2.00 m, as shown in Figure. The child starts from rest at the top. (A) Determine his speed at the bottom, assuming no friction is present. Solution:

  38. (B) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf =3.00 m/s and m = 20.0 kg. Solution:

  39. Example: A block having a mass of 0.80kg is given an initial velocity vA = 1.2m/s to the right and collides with a spring of negligible mass and force constant k = 50N/m, as shown in Figure. (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. Solution:

  40. (B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 0.50. If the speed of the block at the moment it collides with the spring is vA = 1.2 m/s, what is the maximum compression xC in the spring? Solution:

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