Work and kinetic energy. LA info session today at 5pm in UMC235 CAPA homework due tomorrow night.

Transcription

1 Work and kinetic energy LA info session today at 5pm in UMC235 CAPA homework due tomorrow night. 1

2 Work I apply a force of 2N in the x direction to an object that moves 5m in x. How much work have I done on the object? F x (N) x (m) W =! F Δ! r =! F Δ! r cosθ = F(Δr)cosθ Since the displacement and force are in the same direction, simply need to multiple them together. So W = F x Δx = (2N)(5m) = 10 J This is the area underneath the force vs position curve. 2

3 Work by varying force Suppose we have a force that varies in x as shown. How do we determine the work done (for example in going from x=0 to x=5)? F x 2 1 (N) Can divide up path into a bunch of small pieces, calculate the work done in each piece, and add them up. If we take the limit as the size of the piece goes to 0, we get an integral. It is also the area under the curve. x (m) W = x 2 F x dx = area under force vs position curve x 1 3

4 Work by spring force Remember the force by a spring is F = kx. Therefore, the force on a spring when you compress or stretch it is F = kx. What is the work done on a spring when stretching from position x 1 to x 2? x 2 W = F x dx = kx dx = x 1 x 2 x kx2 x 1 x 2 = 1 2 k # x 2 2 $ 2 x 1 % & If we start from the equilibrium point (x 1 =0) then W = 1 2 kx2 kx F x (N) This is the area underneath the curve (a right triangle with sides x and kx). x x (m) 4

5 Even more details about work When working in more than one dimension, the work is further! x modified from 2 r! W = F x dx to the line integral W = F d r! 2! x 1 So at each small displacement dr multiply the displacement by the force acting in that direction. Add up all of these small works to get the total work. r For a car going around a corner, the force causing the radial acceleration is always perpendicular to the velocity, and therefore it is always perpendicular to and will do no work. d! r Since velocity is usually perpendicular to the normal force, most of the time a normal force will not do work either. r 1 5

6 Clicker question 1 A block moves along the x-axis and is acted upon by a variable force as shown in the graph. Between x=0m and x=5 m, how much work is done on the block? A. -1 J B. 0 J C. 3 J D. 4 J E. 5 J Set frequency to BA F x (N) The work done by the force can be found from the area under the curve. W = 2N 2m 1 2 2N 1m = 4 J 1J = J x (m) 6

7 Clicker question 2 A 1 kg mass is moved part way around a square loop as shown. The square is 1 m on a side and the final position of the mass is 0.5 m below its original position. Assume g = 10 m/s 2. What is the work done by the force of gravity during this journey? A. 10 J B. 5 J C. 0 J D. 5 J E. 10 J Set frequency to BA 1.0 m up 0.5 m start finish No work from the horizontal segments since force and displacement are perpendicular Upward segment: W g, up = F! g Δ r! = ( 1 kg) ( 10 m/s 2 )( 0.5 m) = 5 J Downward segment: W W g, down = F! g Δ r! = ( 1 kg) ( 10 m/s 2 )( 1 m) =10 J g, total = Wg, up + Wg, down = 5 J 7

8 Clicker question 2 A 1 kg mass is moved part way around a square loop as shown. The square is 1 m on a side and the final position of the mass is 0.5 m below its original position. Assume g = 10 m/s 2. What is the work done by the force of gravity during this journey? A. 10 J B. 5 J C. 0 J D. 5 J E. 10 J Total work by gravity: Set frequency to BA 1.0 m up 0.5 m start finish Can also think of the 0.5 m upward segment being canceled out by half of the 1 m downward segment leaving just a 0.5 m downward segment W g, total = F! g Δ r! = ( 1 kg) ( 10 m/s 2 )( 0.5 m) = 5 J 8

9 Clicker question 3 Set frequency to BA A 1 kg mass is moved on a totally 1.0 m crazy path that ends up in the same up place as in the previous problem. Is start the work done by gravity during this 0.5 m journey the same or different compared to the previous path? finish A. Same B. Different No work comes from any horizontal motion C. Impossible to tell With the exception of the 0.5 m of difference between the start and end, all other upward and downward motion cancels out Therefore, the work done by gravity is the same! 9

10 Kinetic energy Can consider W = F! Δ r! work = F! done Δ r! by an individual force cosθ = F(Δr)cosθ If you add up the work done by all the forces (or compute the work done by the net force) you get the net work: W net. By Newton s second law, F net = ma = m(dv/dt) so W net = F net dx = Doing this integral, we get: m dv dt dx = mdv dx dt = mv dv W net = v 2 mvdv = 1 v 1 2 mv2 v 1 v 2 = 1 mv mv We define the kinetic energy of an object as K = 1 2 mv2 10

11 Work-energy theorem The work-energy theorem states ΔK =Wnet ΔK = 1 mv 2 2 f 1 mv 2 i2 W =! F Δ! r If the net work is positive, kinetic energy is added to the object and if the net work is negative, kinetic energy is removed from the object. This is always true! But remember the net work means the work done by all forces (or equivalently, the work done by the net force). 11

12 Clicker question 4 Set frequency to BA A hockey puck sliding on a frictionless ice rink at 1 m/s slides onto a rug someone left on the ice. The puck comes to rest after moving 1 m on the carpet. Which forces did work (either positive or negative) while the puck was on the carpet? A. normal force B. gravity C. friction D. more than one of A-C E. None of A-C The normal force and gravity are both perpendicular to the displacement. Therefore they do no work. Kinetic friction force is always opposite the velocity vector so it always does negative work (reducing the kinetic energy). 12