The spring pulls harder, but the block still does not move.

PUM Physics II - Dynamics Lesson 14 Solutions Page 1 of 8 14.1 Observe and Find a Pattern a) The block sits on the table with no scale pulling it. The spring pulls on the block, which does not start moving. The spring pulls harder, but the block still does not move. The spring pulls on the block, and the block is just about to start moving. The spring pulls the block at a slow, constant velocity. F SSonB F SSonB F SSonB F SSonB b) The magnitude of the force that the table exerts in the vertical direction remains the same throughout. However, the force that the table exerts in the horizontal direction increases with the force of the spring scale on the block. It reaches a maximum and then drops to a constant amount once the block starts moving. c) The magnitude of the force that the spring scale has to exert on the block is greater just before the block starts to move. Once it starts moving, the spring scale reading drops. d) The table. e) While at rest, the friction force increases as you increase the amount of force you exert on the object. This reaches a maximum. After this point the object starts moving and the friction force decreases to a steady amount. 14.2 Observe and Find a Pattern Based on the data, the maximum static friction force does not depend on the surface area. It does however, depend on the roughness of the two surfaces.

PUM Physics II - Dynamics Lesson 14 Solutions Page 2 of 8 14.3 Observe and Find a Pattern b) F S max α N surface on object 14.4 Test your idea Students can repeat the experiment from the first activity using their shoe and a level surface. They should increase the amount of mass in the shoe and see if the maximum static friction force increases linearly. The slope of this line should be equal to µ, which they won t know since the coefficient can only be determined experimentally. However, they don t need to know it since they should just be looking for a linear relationship to test the idea. 14.5 Reason a) The maximum static force increases as my neighbor pushes down harder because the normal force of the table on the book increases. It becomes harder to get the book to start moving. Dragging with pinky. Pushing down lightly. Pushing down harder. N TonB N TonB f TonB N TonB F PonB f TonB F NonB F PonB f TonB F NonB F PonB b) The force that Earth exerts on the object always stays constant. It is a function of te object s mass. However, the normal force that the table exerts on the book increases as my neighbor presses down harder. If static friction were dependent on the force of Earth on the book rather than the normal force of the table on the book, then it would not be

PUM Physics II - Dynamics Lesson 14 Solutions Page 3 of 8 any harder to drag the book as my neighbor pressed down harder and harder. This is not what we observe. c) The frictional force between the wall and the book as well as your hand and the book technically prevents the book from falling down. If you do not push hard enough against the book, the maximum static friction is overcome by the force of Earth on the book and kinetic friction instead takes over as the book slides down. f WonB N WonB F HonB 14.6 Design an Experiment Students can produce a graph of the applied force and the friction force vs. time and gradually increase the force until the object starts moving. At this point in the graph there should be a sharp drop in the magnitude of the force applied and the frictional force. Example: The cabinet is 200 kg. Assuming it is on a level surface N SonC = mg = (200 kg)(9.8 m/s 2 ) = 1960 N The maximum force applied is between 588 and 589 N with an uncertainty of about ±1 N. This is equal to the maximum static friction force since this is the point just before the cabinet starts moving. The coefficient of static friction then is: µ s = F s max /N SonC = (588 N)/(1960 N) = 0.3 To determine the coefficient of static friction. Graph the frictional force as the object is sliding. It remains constant at 392 N µ k = F k surface on object /N SonC = (392 N)/(1960 N) = 0.2 These can be double checked against the actual values by clicking on More Controls. They do indeed match meaning that the programmers used Newton s Laws to design their simulation.

PUM Physics II - Dynamics Lesson 14 Solutions Page 4 of 8 14.7 Observe and Represent I ll take my system to be the shoe on my foot. Homework Walking to the right, as I put my foot down, finishing up the last step, the force diagram in the horizontal direction looks like this: F FonS f GonS As I m putting my shoe down, my foot still wants to move forward. However the ground pushes in the opposite direction, bringing my shoe to a stop. The frictional force of the ground on my shoe keeps me from slipping. As I push off this foot the force diagram looks like this: f GonS My foot is pushing to the left on my shoe while the frictional force of the ground on my shoe pushes to the right. This keeps my shoe planted and at rest while the rest of my body moves forward around it. 14.8 Evaluate She could be talking about the internal frictional forces that may slow down the axle and wheel as they spin. However, reducing the frictional force between the tire and ground would not make the car go faster. Without the force of friction between the tire and ground, the tire would not be able to push off and move the car forward, much like how you move when you walk. An easy way to observe this is to put the car on ice. It will not move forward because the tires will spin out, unable to grip. 14.9 Represent and Reason a) Sketch (This is a very odd way to move a desk) + F FonS b) Force Diagram in Horizontal Direction. Vertical forces are assumed to be balanced: F OonD f FonD F JonD F DonD Δv c) a x = [F DonD + (-F JonD ) + (-F OonD ) + (-f k FonD )]/(m D )

PUM Physics II - Dynamics Lesson 14 Solutions Page 5 of 8 d) a x = [F DonD + (-F JonD ) + (-F OonD ) + (-f k FonD )]/(m D ) m D *a x = F DonD + (-F JonD ) + (-F OonD ) + (-f k FonD ) 27 N = F DonD + (-125 N) + (-150 N) + (-200 N) F DonD = 502 N e) The desk is speeding up because there is a net force of 27 N on it. In my sketch that means that the desk is accelerating to the right because there is an unbalanced force. f) The net force would be much greater and the desk would speed up much more rapidly. The force of kinetic friction on the desk would remain constant at -200 N. 14.10 Regular Problem Sketch and Translate v = 97 km/h v = 0 Simplify and Diagram 48 m Since the car is moving as it is sliding to a stop this is definitely kinetic friction that we are dealing with. We ll assume that the car is slowing down with constant acceleration and the road is level. Right is positive. FD: N RonC f k RonC Represent Mathematically v o = 97 km/h = 27 m/s x-x o = 48 m Determine acceleration: v 2 = v o 2 + 2a(x-x o ) 0 = (27 m/s) 2 + 2a(48 m) a = -(729 m 2 /s 2 )/(96 m) = -7.6 m/s 2 F EonC

PUM Physics II - Dynamics Lesson 14 Solutions Page 6 of 8 a = f k RonC /m f k RonC = ma f k RonC = µ k N RonC = µ k mg ma = µ k mg µ k = a/g = (7.6 m/s 2 )/(9.8 m/s 2 ) = 0.8 14.11 Regular Problem Assuming the box is on level ground so that the normal force of the ground on the box is equal to the force of Earth on the box, the maximum static friction force is when: a) f s max = µ s N SonO = µ s mg = (0.70)(50 kg)(10 m/s 2 ) = 350 N FD: F YonB f s max Δv = 0 350 N then is also the minimum amount of force you need to apply to the box so that it overcomes static friction and begins to slide. b) Once the box starts to slide kinetic friction takes over, which is constant assuming N SonO - remains constant. f k = µ k N SonO = µ k mg = (0.50)(50 kg)(10 m/s 2 ) = 250 N FD: F YonB f k Δv If the person is still pushing with 350 N of force. The total force is 350 N 250 N = 100 N. To find the acceleration: a x = F net /m B = (100 N)/(50 kg) = 2 m/s 2

PUM Physics II - Dynamics Lesson 14 Solutions Page 7 of 8 14.12 Reason a) It is the same situation as in 14.5 c) Reason. In the horizontal direction, the leaf is being pushed into the car by the air and the front of the car is exerting a normal force on the leaf. In the vertical direction Earth is pulling down on the leaf, and the front of the car is also exerting a frictional force upwards on the leaf to keep it in place. down. f ConL F AonL N ConL F EonL 14.13 Regular Problem Sketch and Translate v = 20 km/h v = 0 Simplify and Diagram? The car slides to a stop so we are dealing with kinetic friction. We ll assume the car slows down at constant acceleration, the ground is level, and that the only force in the horizontal direction is friction. FD: N RonC f k RonC Represent Mathematically a) m = 1520 kg µ k = 0.40 v o = 20 m/s Determine magnitude of acceleration: f k = µ k N RonC = µ k mg F EonC a x = f k /m = µ k mg/m = µ k g = (0.40)(10 m/s 2 ) = 4 m/s 2

PUM Physics II - Dynamics Lesson 14 Solutions Page 8 of 8 How far does the car go? v 2 = v 2 o + 2aΔx 0 = (20 m/s) 2 + 2(-4 m/s 2 )Δx Δx = -(20 m/s) 2 /(-8 m/s 2 ) = 50 m b) The stopping distance would be the same. Mass doesn t play a role in determining the acceleration and thus the stopping distance. This makes sense. A larger car means a larger normal force and a larger frictional force. However, because the car is more massive you also require a larger force to slow it down at the same rate. Therefore the stopping distance doesn t change because these two factors balance each other out. 14.4 Reason (d) Not enough information to answer. We don t really know for sure. It depends if the grass behaves differently depending on how it is bent over and in what direction ie if the coefficient of static friction is different. We also don t know if the ground is level either. Pushing it down a hill and pulling it up a hill would be different situations.