Relationship between POTENTIAL ENERGY and FORCE

Size: px

Start display at page:

Download "Relationship between POTENTIAL ENERGY and FORCE"

Claude Shields
6 years ago
Views:

1 PH-211 Relationship between POTENTIAL ENERGY and FORCE Knowing F at every place, we defined the corresponding potential energy function U. Here we explore: Given U, how to find F? A. La Rosa = Becomes Becomes

2 Strategy:

4 SUMMARY: CONSERVATIVE FORCE and POTENTIAL ENERGY Conservative forces Relationship between F and U Conservation of the Mechanical Energy

6 Conservative forces Position B Position A

8 For a conservative force There exist a corresponding potential energy U such that,

9 Examples of conservative forces: 1) Gravitational force 2) Spring force

11 Two conservative forces acting at once In this region only the gravitational force acts on the block In this region both the gravitational and spring forces will act on the block

14 METHOD-1 Solution using the concepts of conservation of the mechanical energy Velocity of the block just before it hits the spring The block compresses the spring 12 cm At z=0

15 METHOD-2 Solution using the work / kinetic-energy theorem ΔK = W Work on the block by the gravitational force Work on the block by the spring force W if = Kg Kg

17 Example. (μ k = 0.2), compression Expected answers: a) m, b) m, c), 1.54m SOLUTION STRATEGY : Let's solve the problem using the theorem ΔK = W

18 First, we derive general expressions to be used in the regions z >z o and z< z o respectively For z > z o z v z o ΔK = K(z o ) - K(z) W total = W normal + W gravitational W total = 0 - mg(z o -z) We choose z o as the height at which the spring is neither elongated nor compressed. mg N Notice that in this regime of motion, we are taking z o as the final position and z as the initial position ΔK = W implies, K(z o ) - K(z) = - mg(z o -z) K(z o ) + m g z o = K(z) + m g z (1)

19 For z < z o K = W total ( work done by all the forces acting on the mass m) K = W normal-force + W spring-force + W gravity-force = 0 - (1/2) (s s o ) 2 - mg (z z o ) Apparently we have two variables z and s. However we realize, sin = (z z o ) / (s s o ) K = - (1/2) (s s o ) 2 - mg (s s o ) sin

S o ) sin(30o) ~ 40 Joules Motion from B to C Using (2): ΔK = - (1/2) (s - s o ) 2 - mg (s - s o ) sin θ Identifying s with s C, and s o with the coordinate of point B: Notice s C < s o and d=s o -

20 Now let's put some numbers. z A z o z C S A We choose z o as the height at which the spring is neither elongated nor compressed. S o S C S A - S o = 4m S o - S C = d =? Motion from A to B Using (1): K(z) + m g z = K(z o ) + m g z o Identifying z with z A, and z o with the coordinate of point B: = 0 K A + mg z A = K AB + mg z B z A K B = mg (z A - z o ) = mg (S A - S o ) sin(30o) ~ 40 Joules Motion from B to C Using (2): ΔK = - (1/2) (s - s o ) 2 - mg (s - s o ) sin θ Identifying s with s C, and s o with the coordinate of point B: Notice s C < s o and d=s o - s C ΔK = K C - K B = - (1/2)k (s C - s o ) 2 - mg (z C - z o ) 0-40 J = - (1/2) k d d 50 d 2-10 d - 40 = 0 d = 1 meter We have used g ~ 10 m/s 2 (just to simplify the calculations) (s C - s o ) 2 - mg (s C - s o ) sin30 o

21 Let's solve now the problem for the case in which there is friction For z > z o z s v First, we derive general expressions to be used in the regions z >zo and z< zo respectively z o s o We choose z o as the height at which the spring is neither elongated nor compressed. ΔK = K(z o ) - K(z) W total = W normal + W gravitational + W friction F f N W total = 0 - mg(z o -z) + F f (s o -s) mg Notice that in this regime of motion, we are taking z o as the final position and z as the initial position ΔK = W implies, K(z o ) - K(z) = - mg(z o -z) + F f (s o -s) [K(z o ) + m g z o ] - [ K(z) + m g z ] = F f (s o -s) (3)

However we realize, sin = (z z o ) / (s s o ) + F f (s -s o ) K = - (1/2) (s s o ) 2 - mg (s s o ) sin + F f (s -s o ) K(z) -

22 For z < z o K = W total ( work done by all the forces acting on the mass m) F f θ K = W normal-force + W spring-force + W gravity-force + W friction = 0 - (1/2) (s s o ) 2 - mg (z z o ) Apparently we have two variables z and s. However we realize, sin = (z z o ) / (s s o ) + F f (s -s o ) K = - (1/2) (s s o ) 2 - mg (s s o ) sin + F f (s -s o ) K(z) - K(z o ) = - (1/2) k(s - s o ) 2 - mg (s - s o ) sin θ + F f (s -s o ) [K(z) + (1/2) k(s - s o ) 2 +mgs Sin θ] - [K(z o ) + mg s o sin θ] = F f (s -s o ) (4)

23 Now let's put some numbers. We choose z o as the height at which the spring is neither elongated nor compressed. Motion from A to B Using (3): [K(z o ) + m g z o ] - [ K(z) + m g z ] = F f (s o -s) z A Identifying z with z A, and z o with the coordinate of point B: [K B + m g z o ] - [ K(z A ) + m g z A ] = F f (s o -s A ) [K B + m g z o ] - [ 0 + m g z A ] = F f (-4 meters ) K B = m g (z A -z o ) + F f (-4 meters ) K B = m g (s A -s o )sin θ + (0.2) mg cos(θ) (- 4 ) (s C - s o ) 2 K B = m g (4 )sin θ + (0.2) mg cos(θ) (- 4 ) K B = 40 + (0.2) mg( 0.86 ) (- 4 ) K B = Joules

24 Motion from B to C Using expression (4) [K(z)+ (1/2) k(s - s o ) 2 +mg s Sin ] - [K(z o ) + mg s o Sin ] = = F f (s -s o ) Identifying s with s C, and s o with the coordinate of point B: [KC + (1/2) k(s C - s o ) 2 +mg s C Sin ] - [K B + mg s o Sin ] = = F f (s -s o ) [ 0 + (1/2) k d 2 + mg s C Sin ] - [K B + mg s o Sin ] = = (0.2) mg cos( ) (- d ) (1/2) k d 2 + mg (s C -s o ) Sin - K B = (0.2) mg cos( ) (- d ) (1/2) k d 2 + mg (- d ) Sin - KB = (0.2) mg cos( ) (- d ) K= 100 N/m, = 30º 50 d d - KB = (0.2) mg cos( ) (- d )

25 50 d 2 + [ (0.2) mg cos( ) - 10 ] d - KB = 0 50 d 2 + [ (0.2) mg cos( ) - 10 ] d = 0 50 d 2 + [ ] d = 0 50 d 2 + [ ] d = 0 50 d 2 - [ 6.53 ] d = 0 d = 0.70 meters

28 Example Equilibrium position Solution a)

29 b) E o = m * (0) 2 = = 1.25 J E o

32 Exercise: Question #4, Chapter 8, (page 188) The figure gives the potential-energy function of a particle. What value must be the mechanical energy Emec of the particle not to exceed if the particle is to be b) trapped in the potential well at the left, c) trapped in the potential well at the right, and d)able to move between the two potential wells but not to the right of point H.

34 Exercise: Question #8, Chapter 7, (page 159)

Physics 2414 Group Exercise 8. Conservation of Energy

Physics 2414 Group Exercise 8. Conservation of Energy Physics 244 Group Exercise 8 Name : OUID : Name 2: OUID 2: Name 3: OUID 3: Name 4: OUID 4: Section Number: Solutions Solutions Conservation of Energy A mass m moves from point i to point f under the action