Soru
Consider the system shown in the figure (Exgure 1) Block A has weight w_(A) and block B has weight wB Once block B is set into downward motion it descends at a constant speed Assume that the mass and friction of the pulloy are negligible Part A Calculate the coefficient of kinetic friction is between block A and the table top Express your answer in terms of some or all of the variables w_(A) subscript A), wp, and g (the acceleration due to gravity). View Available Hint(s) square sqrt [square ](square ) Asum phi ? x^a x_(b) √x Vx x x xcdot 10^n D mu =(w_(b))/(w_(a))
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Usta · 5 yıl öğretmeni
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To calculate the coefficient of kinetic friction between block A and the table top, we need to consider the forces acting on block A.The force of kinetic friction (F_k) is given by the equation:F_k = μ * Nwhere μ is the coefficient of kinetic friction and N is the normal force acting on block A.Since block A is in equilibrium and moving at a constant speed, the net force acting on it is zero. Therefore, the force of kinetic friction is equal to the force of gravity acting on block B.The force of gravity acting on block B is given by:F_g = w_B * gwhere w_B is the weight of block B and g is the acceleration due to gravity.Since the mass and friction of the pulley are negligible, the force of kinetic friction is equal to the force of gravity acting on block B:F_k = F_gSubstituting the values, we have:μ * N = w_B * gThe normal force (N) acting on block A is equal to the weight of block A:N = w_A * gSubstituting this value into the equation, we get:μ * w_A * g = w_B * gSimplifying the equation, we find:μ = w_B / w_ATherefore, the coefficient of kinetic friction between block A and the table top is given by:μ = w_B / w_AThis expression represents the coefficient of kinetic friction in terms of the weights of blocks A and B, and the acceleration due to gravity.