![]() Add masses to the block in 200g increments up to 1200g. Add weight to the weight ngar (m) until the block slides with constant speed when started in motion with a light tap. Touching the sliding surface with sticky or greasy hands will change the coefficient of friction. The wooden blocks M should only be held by the sides. Make sure the pulley is adjusted so the string is arallel to the sliding surface. Notice that the pulley is adjustable in height. This will insure the grain of the wood is the same direction as in part 2 of this experiment. The pulley should be attached to the end of the block without the hole through it. Run a light string from the block M, over the pulley, and attach a weight hangar. Place the plane board flat on the table, attach a pulley to one end and place the block on the plane. 0) to derive the equation:ĮXPERIMENTAL PROCEDURE: Part 1: Take the mass of the block (M=509 g). A) Part 1: The student should derive the equation: (1) MXM = m (at a constant speed) B) Part 2: The student shall derive the equation for the angle of repose, Or, for kinetic friction (the angle for which the block slides down the plane at constant speed). When fx is plotted on the y-axis and Fy is plotted on the x-axis, the slope of the straight line is M. ![]() The net force acting on the skier in her direction of motion is on the box is 445.2 N.The coefficient of kinetic friction is define by the equation: MX F re-arranging the equation gives fa- MF. If we define the positive direction as down the hill, which is the skier's direction of motion, the net force is: The two forces to consider are the force directed down the hill and the force of kinetic friction directed up the hill. The net force acting on the skier is the sum of forces. The force that moves the skier down the hill is the remaining component of the force of gravity, F θ = mg sinθ. The force of kinetic friction opposes the motion of the skier. Using this, the formula can be used to find the magnitude of the force of kinetic friction:į k = (0.0800)(55.00 kg)(9.80 m/s 2)cos(60°) The normal force is η = mg cosθ, and the remaining force component is F = mg sinθ. The normal force is the component that is perpendicular to the angled surface, and the remaining force is parallel to the angled surface. On a surface that is at an angle relative to the horizontal axis, The total force due to gravity, F = mg, must be broken in to components. What is the magnitude of the force of kinetic friction, and what is the net force along the skier's direction of motion?Īnswer: On a flat surface, the normal force on an object is η = mg. The hill is at a 60.0° angle from the horizontal. The coefficient of kinetic friction between the skis and the snow is μ k = 0.0800. The net force acting on the box is 17.8 N forward.Ģ) A woman is skiing straight down a snow-covered hill. If we define "forward" as the positive direction, the net force is: The two forces to consider are the force of kinetic friction acting in the opposite direction as the box's motion, and the force exerted by the worker, which is 400 N forward. The net force acting on the box is the sum of forces. The force of kinetic friction acting in the opposite direction as the motion of the box is 382.2 N. Using this, the formula can be used to find the force of friction: What is the magnitude of the force of friction, and what is the net force moving the box?Īnswer: On a flat surface, the normal force on an object is η = mg. The box has a mass of 75.0 kg, and the worker is exerting a force of 400.0 Nforward. The coefficient of kinetic friction is μ k = 0.520. 1) A worker in a stock room pushes a large cardboard box across the floor.
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