Consider what happens if the block is suddenly disconnected from the spring, as shown below. Choose the correct equation for calculating the coefficient of kinetic friction (lk) if it takes time t for the block to travel a displacement of Ax down the inclined surface. Note that this displacement is negative in the given rotated coordinate system, and that the acceleration of the block is constant so the motion is described by the four kinematic equations. Hint: remember that Ar = r TO
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- A desperate hiker has to think fast to help his friend who has fallen below him. Quickly, he ties a rope to a rock of mass ma and makes his way over the ledge (see the figure). If the coefficient of maximum static friction between the rock and the ground is µ, and the mass of the hiker is mâ, what is the maximum mass of the friend, mc, that the rock can hold so the hikers can then make their way up over the ledge? Assume the rope is parallel to the ground and the point where the rope passes over the ledge is frictionless. mc = JuTwo blocks of masses m, = 6.0 kg and m2 = 12 kg are connected with a string that passes over a very light pulley (Figure 1). Friction in the pulley can be ignored. Block 1 is resting on a rough table and block 2 is hanging over the edge. The coefficlent of friction between the block 1 and the table is 0.70 (assume static and kinetic friction have the same value). Block 1 is also connected to a spring with a constant 300 N/m. In the initial state, the spring is relaxed as a person is holding block 2, but the string is still taut. When block 2 is released, it moves down for a distance d until it stops (the final state). Figure < 1 of 1Needs Complete typed solution with 100 % accuracy.
- a tilted applied force, which requires that we work with components to find a frictional force. The main challenge is to sort out all the components. a force of magnitude F = 12.0 N applied to an 8.00 kg block at a downward angle of u= 30.0. The coefficient of static friction between block and floor is ms = 0.700; the coefficient of kinetic friction is mk = 0.400. Does the block begin to slide or does it remain stationary? What is the magnitude of the frictional force on the block?Be sure all quantities are expressed in standard units before calculating anything. Thank you :)I am trying to solve this problem but the equation I came up with dosn't work and I am at a loss with what I might be missing for this so can you please help me?
- In the pulley system shown in the figure, one end of a string is connected to the end of a rod, which can rotate about a fixed pin. The length of the string, L, is constant. The rod has a constant length, h, which is equal to the fixed height h in the figure. The other end of the string is attached to a block that can move in the vertical direction. Find the acceleration of block Aa® with respect to a fixed frame. h PLearning Goal: To be able to set up and analyze the free-body diagrams and equations of motion for a system of particles. Consider the mass and pulley system shown. Mass m₁ = 35 kg and mass m₂ = 12 kg. The angle of the inclined plane is given, and the coefficient of kinetic friction between mass m₂ and the inclined plane is 0.19. Assume the pulleys are massless and frictionless.(Figure 1) Figure M₂ M₁ < 1 of 1 ▼ What is the acceleration of mass me on the inclined plane? Take positive acceleration to be up the ramp. Express your answer to three significant figures and include the appropriate units. ▸ View Available Hint(s) Submit v(45)= º Submit μA Value Part B - Finding the speed of the mass moving up the ramp after a given time If the system is released from rest, what is the speed of mass m₂ after 4 s? Express your answer to three significant figures and include the appropriate units. ▸ View Available Hint(s) Submit ! HÅ Value Units 2= Value µÅ A Part C- Finding the distance moved by…Calculate the magnitude of the normal force on a 17.7 kg block in the following circumstances. (Enter your answers in N.) The block is on a level surface and a force of 165 N is exerted on it at an angle of 40.8° above the horizontal. I don't know how to do this.