Potential Energy and Conservative Forces Chapter (8) Potential Energy and Conservative Forces

There are two types of forces: conservative (gravity, spring force) All microscopic forces are conservative: Gravity, Electro-Magnetism, Weak Nuclear Force, Strong Nuclear Force nonconservative (friction, tension) Macroscopic forces are non-conservative,

Conservative Forces  work done depends only on ri and rf A force is conservative if the work it does on an object moving between two points is independent of the path taken.  work done depends only on ri and rf

 If an object moves in a closed path (ri = rf) then total work done by the force is zero.

Nonconservative Forces  work done by the force depends on the path  non-conservative forces dissipate energy

Work Done by Conservative Forces Potential Energy: Energy associated with the position of an object. For example: When you lift a ball a distance y, gravity does negative work on the ball. This work can be recovered as kinetic energy if we let the ball fall. The energy that was “stored” in the ball is potential energy. Wc = -DU =-[Ufinal – Uinitial] Wc = work done by a conservative force DU = change in 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)

Spring Potential Energy Uf – Ui = - [Work done by spring on mass] Mass m starts at x=0 (Ui =0) and moves until spring is stretched to position x. WorkSpring = - ½ kx2 U(x) – 0 = - (-1/2 kx2) USpring(x) = ½ kx2 x = displacement from equilibrium position F=-kx x Area in triangle = -kx times increment in x = Work done by spring

Conservation of Energy Energy is neither created nor destroyed The energy of an isolated system of objects remains constant.

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

Problem: find the particle’s speed at points B, where the particle released from point A and slides on the frictionless track Problem: Find the distance x.

Problem: A 0. 2-kg pendulum bob is swinging back and forth Problem: A 0.2-kg pendulum bob is swinging back and forth. If the speed of the bob at its lowest point is 0.65 m/s, how high does the bob go above its minimum height? Problem: Two objects are connected by a light string passing over a light frictionless pulley as shown in Figure. The object of mass 5.00 kg is released from rest. Using the principle of conservation of energy, (a) determine the speed of the 3.00-kg object just as the 5.00-kg object hits the ground. (b) Find the maximum height to which the 3.00-kg object rises.

Problem: An object of mass m starts from rest and slides a distance d down a frictionless incline of angle . While sliding, it contacts an unstressed spring of negligible mass as shown in Figure. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring.

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 = 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?

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 +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.

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.