2.0 kg push 45° Problem 4: (a) The 2 kg wood box in the Figure above on the left slides up a vertical wood wall while you push on it at a 45° angle. The coefficient of kinetic friction for wood on wood is = 0.2. What magnitude force Fpush should you apply to cause the box to slide up at a constant speed? Recall that constant speed means a = 0 m/sec2. Answer: 34.6 N.

You pull on a crate using a rope as image, except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of the crate is 325 N, and the coefficient of kinetic friction between the crate and the floor is 0.230. What must be the tension in the rope to make the crate move at a constant velocity? What is the normal force that the floor exerts on the crate?

A carnival ride has people stand inside a vertical cylinder with their backs to the wall. The cylinder starts spinning and the riders find that they are “stuck” to the wall and don’t slide down, even if the floor is removed.  The ride has a radius of r. The person has a mass of m and is moving with a constant speed of v. The coefficient of static friction between the person and the wall is μs , and kinetic friction μk. The person is only touching the wall, not touching the floor.   What is the magnitude of the acceleration of the person? In what direction does it point?   The speed is constant. Why is the acceleration not zero? Briefly explain. No equations!

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You construct a version of the cart and bucket, but with a slope whose angle can be adjusted. You use a cart of mass 167 kg and a bucket of mass 57.0 kg. The cable has negligible mass, and there is no friction. What must be the angle of the slope so that the cart moves downhill at a constant speed and the bucket moves upward at the same constant speed? With this choice of angle, what will be the tension in the cable?

The figure shows a graph of force against distance for a locomotive pushing a carriage along a train track. x (m) Note: Please enter all your answers in this question to full precision. For the case where x:= 330 m x,= 750 m x, =1260 m and Fmax 31000 N, calculate the work done by the locomotive in megajoules for: (a) the part of the journey where the force is increasing (i.e. the first stage): MJ (b) the part of the journey where the force is constant (i.e. the second stage): MJ (c) the part of the journey where the force is decreasing (l.e. the third stage): MJ (d) Calculate the total work done by the locomotive: MJ

Solve it.

3.2 Sliding blocks with friction* Mass MA = 4kg rests on top of mass Mg = 5 kg that rests on a fric- tionless table. The coefficient of friction between the two blocks is such that the blocks just start to slip when the horizontal force F applied to the lower block is 27 N. Suppose that now a horizontal force is applied to the upper block. What is its maximum value for the blocks to slide without slipping relative to each other?

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A uniform ladder 8 meters long and weighing 350 N rest against a smooth vertical wall at an angle of 30° to the wall. A 700-N man stands 6 meters up from the bottom of the ladder. Find the horizontal force necessary at the base to keep the ladder from slipping.   Ps: Include Given, Free-Body Diagrams and Formula