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Work, Energy, Power and Momentum

Work, Energy, Power and Momentum. Work.

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Work, Energy, Power and Momentum

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  1. Work, Energy, Power and Momentum

  2. Work • We know that when a force is applied to an object, the object will be displaced some distance, x, in the direction of the force. If the force pulling the object is not applied directly along the path of motion, then we must find the component of the force that is along the path of motion. (A rope being used to pull a crate at an angle above the horizontal).

  3. Work is defined as the component of applied force times distance over-which that force is applied. • W = (F cos Θ)(x) • When the force is directly along the direction of motion, Θ = 0 and cos Θ = 1 • The angle Θ is between the force and displacement vectors. • Work is scalar, not a vector. • W = N x m = (kg●m/s2)(m) = kg●m2/s2 = joule (J)

  4. A worker pulls a crate with a force of 250.0 N for a distance of 3.20 m applying the force parallel to the floor and in the direction of motion. Find the work done by the worker on the crate, Wwc. W = (F cos Θ)(x) = (250N )(3.20 m)= 800. j If the force in the above problem were applied with a rope at an angle of 35.0° above the horizontal, find the Wwc.

  5. Only the component in direction of force causes work

  6. Each force that acts on a body has the potential to do work, and to determine the net work, Wnet, we must take the total of all the individual amounts of work. • Forces applied perpendicular to the displacement of an object do W = 0 • Generally when a force applied is opposite the direction of the displacement of the object, the work done by that force will be Negative. Friction would be an example of this, as would a force slowing down a moving object.

  7. A 20 kg crate is to be moved at constant velocity, 7.0 m across a warehouse floor by applying a force parallel to the floor in the direction of motion. The coefficient of friction is 0.20. Determine the amount of work done by friction, Wfc the amount of work done by the worker Wwc, and the net work, Wnet, if the crate is (a) moved uniformly, or (b) if the worker applies a 100 N force parallel to the floor.

  8. Wfk = fk (cos180)d = – µmg(d) = – (0.2)(20kg)(10 m/s2)(7.0m) = – 280 J Constant Velocity so fk must equal Fwc = 40N Wwc = F(cos 0)d = 40 (1)(7.0m) = + 280 J Wnet = Wfk + Wwc = – 280 J + 280 J = 0 What if the worker pulled with 100 N? Wwc = 100 (1) (7.0m ) = 700 J and Wnet = – 280 J + 700 J = 420 J

  9. A worker pulls a 40 kg crate as shown above. µk = 0.550. If he moves the crate with constant velocity a distance of 7.00 m, how much work does he do? (Need to find Fapplied first)

  10. Σ Fx = max = 0 = F cos30 – fk = F cos30 – µN Σ Fy = may = 0 = N + F sin 30 – mg Solve simultaneously: N = mg – F sin30 Σ Fx = F cos 30 - µ (mg – F sin 30) = F cos30 – µmg + µF sin 30 µmg = F (cos 30 + µF sin 30) µmg______ = F = 189 N cos 30 + µF sin 30 Wwc = F cos 30 (d) = 189 (cos 30)(7.0) = 1145J

  11. A 0.75 kg block slides with uniform velocity down a 20º inclined plane. Find the work done by friction, Wf on the block as it slides down total length of the plane. What is the net work done on the block? Discuss the net work done if the angle of the plane were adjusted so that the block accelerates down the plane.

  12. Wf = fk (cos 180º)d = -fkd = -µNd (180º opposite displace = – 1 • d can be found from trig d = L/cos Θ • fk must be equal to the component of weight parallel to the plane, or mg sinΘ • By substitution: Wf = – fkd = – mg sin Θ (L/cos Θ)= – mg L tan Θ Wf = – (0.75kg)(9.8 m/s2)(1.2 m)(0.364) = -3.2 J Wg = mg sinΘ = (0.75kg)(9.8 m/s2)(.342) = 3.2 J Wnet = Wf + Wg = – 3.2 J + 3.2 J = 0

  13. Energy • Varying forms of energy make a concise definition difficult. Basically, energy is the ability to do work. All forms of energy can be converted into work. Most commonly, objects can possess energy of motion which can be converted into work upon impact with another object.

  14. Energy of motion, kinetic energy, can be calculated by using KE = ½ mv2. • This definition is derived from: w = (F)(x), F= (m)(a), a =V/t and x=1/2 at2 from kinematics. w= (F)(x) = [ma][ ½ at2] = [(m)(V/t)][ ½ (V/t)t2] = ½ m (V2/t2)t2 =1/2 mV2 = kg m2/s2 = J This is specifically for translational kinetic energy.

  15. To find out the amount of work done on an object, you must know the change in kinetic energy of the object. • ΔKE = ½ mVf2- ½ mV02this is known as the work energy theorum. • Find the KE of a 0.17 kg ball thrown at a velocity of 36 m/s. KE = ½ mV2 = ½(0.17kg)(36m/s)2 = 110 j

  16. Find the velocity of a 1250 kg auto if its initial KE is 15,000 j. • 15,000 = ½ (1250)(v)2 • v = 4.90 m/s • How much work is required to increase the velocity of a 50.0 kg moped from 15km/hr to 22km/hr. w = ½ 50.0kg(6.11m/s)2 – ½ 50.0kg (4.17m/s)2 w = 933 j – 435j w = 498 j

  17. Gravitational Potential • When work is done on an object vertically, the amount of work done moving the object up is stored in the object as potential energy while the object is waiting to be released. We call this energy due to height, gravitational potential energy as Fw is the force which is acting on the object. • U = Fwh = (m)(g)(h) • U is relative to some lower surface, often the ground.

  18. We have all heard of the law of conservation of energy which will apply in mechanics as well. Etot = KE + PE. This is true at any time in the mechanical system. • A 1kg book sits on a desk 1.0 m above the ground. What is its KE when it hits the ground? • PEtop + KEtop = Etot = PEbot + KEbot • mgh + ½ mV2 = mgh + ½ mV2 • 9.8 j + 0 = 0 + x • 9.8 j = KEbot

  19. We could also find the velocity of the book before it hits the ground. • 9.8 j = ½ mV2 = ½ (1)(V2) = 4.43 m/s • For any mechanical case involving no friction and vertical motion: • ½ mV12 + mgh1 = ½ mV22 + mgh2 • This equation can be solved for any of the variables, and notice that we do not really need mass.

  20. Power • In science, the term power refers to the time rate of work, P = w/t. This would be J/s which are known as watts. P = F v too! • Vertically w = mgh, so power is mgh/t. • Climbing stairs involves the same amount of work regardless of the time, but the power will vary greatly due to the time involved.

  21. Power is also the product of force x velocity. • (N)(m/s) = J/s = Watts

  22. Power can also be calculated as the product of force x velocity. (

  23. Linear Momentum • Another quantity that relates mass and velocity is called momentum. • Momentum, ρ, is the product of an objects mass and velocity. ρ = (m)(V) • When two objects collide, there are two principles which must be considered. • The total momentum must be conserved • The objects act with impulsive, short lived, forces.

  24. Impulse is the change in momentum over time, or more usefully, Δρ = (F)(t). • m(Vf- V0) = (F)(t) • Used to calculate the extraordinary forces encountered during collisions.

  25. Conservation of Momentum • Provided there are no external forces acting on a system, the total momentum before a collision must equal the total momentum after the collision. • For elastic collisions involving two bodies: ρbefore = ρafter m1V1 + m2V2 = m1V1’ + m2V2’ where V1’ and V2’are the velocities after the collision.

  26. A 0.50 kg ball moves east with a velocity of 0.75 m/s and collides with a stationary ball of mass 0.65 kg. After the collision the 0.50 kg ball has a velocity of 0.15 m/s. Find the velocity of the bigger ball. m1V1 + m2V2 = m1V1’ + m2V2’ (0.50kg)(0.75m/s)+(0.65kg)(0m/s)= (0.50kg)(0.15m/s)+(0.65kg)V2’ V2’= 0.46 m/s

  27. A ball of mass 1.57 kg is rolling west with a velocity of 2.00 m/s. it is struck from behind by a ball with a velocity of 3.50 m/s. If the velocity of ball 1 is 2.25m/s and ball 2 has a velocity of 1.95 m/s after the collision, find the mass of ball 2. m1V1 + m2V2 = m1V1’ + m2V2’ (1.57kg)(2.00)+(M2)(3.50) = (1.57 kg)(2.25)+(M2)(1.95) M2 = 0.39/1.55 M2 = 0.252 kg

  28. For inelastic collisions, where the bodies remain connected after the collision. • Momentum is still conserved in inelastic collisions, however, there is only one object moving after the collision. • m1V1 + m2V2 = (m1 + m2)V’ • An ice skater with mass 75 kg is standing motionless when he is struck from behind by an 85.0 kg skater moving at 1.75 m/s. What is the final velocity of the pair? • (75.0 )(0) + (85)(1.75) = 160V’ • V’ = 0.930 m/s

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