A small box of mass m1 is sitting on a board of mass m2 and length L. The coefficient of kinetic friction between the board and the horizontal floor is μk . The coefficient of static friction between the board and the box is μs. At this moment, the board slides to the right along the surface, while the box does not slide on the board. What is the maximum magnitude of the force F pulling the board so that the box does not slip on the board? * the answer key is (Us + Uk)(m1 + m2)g, I don't know how to get here.

A small box of mass m1 is sitting on a board of mass m2 and length L. The coefficient of kinetic friction between the board and the horizontal floor is μk . The coefficient of static friction between the board and the box is μs.At this moment, the board slides to the right along the surface, while the box does not slide on the board. What is the maximum magnitude of the force F pulling the board so that the box does not slip on the board? * the answer key is (Us+Uk)(m1+m2)g, I don't know how to get here.

Problem 2: Princess Peach is competing in a Mario Kart race on Rainbow Road. Rainbow Road has no rails, so Peach has to be careful taking turns to make sure she doesn't fall off. There is a notoriously difficult turn on the track made up of a portion of a circle 120m in diameter. The road is angled at 30°. The track has a friction coefficient of μ = .40. Given that Peach and the car weigh 800 kg. a) Draw a free-body diagram for the car on the road. b) What is the magnitude of the normal force, and what angle does it make with the horizontal? c) Repeat part b for the the friction force d) How fast can Peach take the turn without flying or falling off the track?

Problem: A girl weighing 62 kg resting on a horizontal surface is having 608.22 N force as shown in the photo. Determine the friction force between the block and the surface. Assume that the coefficient of the static friction between the girl and the surface is 0.05. Solution A. Given w= 62 kg F= 608.22 N Hs= 0.05 B. Free-Body Diagram 608-22 N Quote FN W=62k9 F

Multiple part question: A worker in a factory slides a heavy box up a ramp into a truck. The box has a mass of 84 kg, and the ramp is 3 m long and makes an angle of 34° with the horizontal. The coefficient of friction between the box and ramp is 0.6. What is the weight of the box? What is the normal force acting on the box? What is the force of friction acting on the box?   Assuming the worker pushes the box up the ramp with a constant velocity, how much work does the factory worker do on the box in order to get it into the truck?   How much gravitational potential energy does the box have once it reaches the top of the ramp?

1 This loudspeaker has a weight of 70 Newtons and is being held up by two cables. Both cables are at an angle of 25 degrees. This means that they both have the same tension, T. What is the vertical component of the tension in each cable? Ty %D unit What is the horizontal component of the tension in each cable? Tx unit What is the total tension in each cable? T = unit %3D

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Suppose you use an ideal pulley of the type shown in image (a) in the figure below to support a car engine of mass 145 kg. MA-2 MA-3 MA-4 Ο (b) (c) (a) What would be the tension in newtons in the rope? N (b) What force in newtons must the ceiling supply, assuming you pull straight down on the rope? Neglect the pulley system's mass. N

GOAL Apply the concept of static friction to an object resting on an incline. PROBLEM Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is ng sin e 0.350, what maximum angle can the ramp make with the horizontal mg cos e e before the block starts to slip down? STRATEGY This is an application of Newton's second law involving an object in equilibrium. Choose tilted coordinates, as in the figure. Use the fact that the block is just about to slip when the force of static friction takes its maximum value, f. = 4n. SOLUTION Write Newton's laws for a static system (1) EF = mg sin e - u̟n = 0 in component form. The gravity force (2) EF = n - mg cos e = 0 has two components. Rearrange Equation (2) to get an n = mg cos 8 expression for the normal force n. Substitute the expression for n into EF, = mg sin 8 - H mgcos 0 = 0 → tan 8 =u. Equation (1) and solve for tan 0. Apply the inverse tangent function to get the answer.…

A 40 N box is resting on a table when a 20 N force is applied to it horizontally. The coefficients of static and kinetic friction are 0.3 and 0.1, respectively. Which of the following best describes the resulting motion of the box?    The motion of the box cannot be determined with the given information.      The box accelerates in the direction of the applied force.      The box remains at rest.      The box moves with a constant velocity in the direction of the applied force.