# Work Done by Non-conservative Forces

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

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

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.

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

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.

(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.

Example: A block having a mass of 0
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.

(B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 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?

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?

Problem: A 1300-kg car drives up a 17. 0-m hill
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  105 J. Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill.

Problem :A skateboard track has the form of a circular arc with a 4
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?

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.

Chapter (9) Review

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?

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

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

Mechanical Energy (Conservative Forces)
Mechanical energy E 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 constant if the objects interact only through conservative forces: E = constant Ef = Ei  Uf + Kf = Ui+ Ki DU + DK = DE = 0

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.

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

Example: A particle of mass m = 5
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.

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

For the car rolling up the incline

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.

Example (1) :A 100g car is taken from the base to the top of a 10
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.

1-

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

Since the path doesn’t matter let’s go straight upward
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,

Example : 00 A pumpkin (نبات القرع) falls 62. 0m to the ground
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.

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

Spring Potential Energy:

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

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 :

Now use the Work-Energy Theorem:

How far does the bicycle travel if it takes 4.0 s to come to rest?
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 = kg m 2 /s 2 = J

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= = 24 m Braking force Wnet = Fnet d = FFriction d FFriction = Wnet /d = (-5400)/(24) = -225 N

7. 7: A 10. 0kg block slides down a 3. 00m high frictionless ramp
7.7: A 10.0kg block slides down a 3.00m high frictionless ramp. Then it skids 6.00m along 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 the coefficient of friction between the block and the rough surface.

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:

(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:

Example: A block having a mass of 0
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:

(B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 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:

Solve the equation to find x

Example: A 1300 kg car drives up a 17. 0 m hill
Example: 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  105 J. Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill. Solution:

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.

Example: A toy car completes one lap around a circular track (a distance of 200m) in 25s. (a) What is the average speed? (b) If the mass of the car is 1.5kg, what is the magnitude of the centripetal force that keep it in a circle? Solution: (a) the average speed      v = 200/25 = 8 m/s (b) the magnitude of the centripetal force = mv2/r

Example: A particle moves in a circular path 0
Example: A particle moves in a circular path 0.4m in radius with constant speed.  If the particle makes five revolution in each second of its motion, find (a) the speed of the particle and (b) its acceleration Solution: (a)    Since r =0.4m, the particle travels a distance 0f 2 r = 2.51m in each revolution. Therefore, it travels a distance of 12.57m in each second (since it makes 5 rev. in the second). v = 12.57m/1sec  =  12.6 m/s (b)

Example: A train slows down as it rounds a sharp horizontal turn, slowing from 90km/h to 50km/h in the 15s that it takes to round the bend.  The radius of the curve is 150m.  Compute the acceleration at the train. Solution:

Example: A 1500-kg car accelerates uniformly from rest to a speed of 10 m/s in 3s. Find (a) the work done on the car in this time, (b) the average power delivered the engine in the first 3s, and Solution: a) b)

Example: When a 4kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5cm. If the 4kg mass is removed, (a) how far will the spring stretch if a 1.5kg mass is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.0 cm from its unstretched position? Solution: a) b)

Example: A single conservative force Fx = (2x + 4) N acts on a 5kg particle, where x is in m. As the particle moves along the x axis from x = 1 m to x = 5 m, calculate (a) the work done by this force, (b) the change in the potential energy of the particle, and (c) its kinetic energy at x = 5 m if its speed at x =1m is 3 m/s. Solution: a) b) c)

Example: Use conservation of energy to determine the final speed of a mass of 5.0kg attached to a light cord over a massless, frictionless pulley and attached to another mass of 3.5 kg when the 5.0 kg mass has fallen (starting from rest) a distance of 2.5 m as shown in Figure Solution: Substitute and find v

Example: A 5kg block is set into motion up an inclined plane as in Figure with an initial speed of 8 m/s.  The block comes to rest after travelling 3 m along the plane, as shown in the diagram.  The plane is inclined at an angle of 30' to the horizontal. (a) Determine the change in kinetic energy. (b) Determine the change in potential energy. (c) Determine the frictional force on the block. (d) What is the coefficient of kinetic friction? Solution:

a) b) c) d)

Example: A block with a mass of 3 kg starts at a height h = 60 cm on a plane with an inclination angle of 30', as shown in Figure. Upon reaching the bottom of the ramp, the block slides along a horizontal surface. If the coefficient of friction on both surfaces is ,Uk = 0.20, how far will the block slide on the horizontal surface before coming to rest? Solution:

And find v And find s

Example: Initially sliding with a speed of 1. 7 m/s, a 1
Example: Initially sliding with a speed of 1.7 m/s, a 1.7 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring? Solution:

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