標題:

kinetic friction

發問:

Suppose the coefficient of kinetic friction between m(A) and the plane in the figure is u_k = 0.10, and that m(A) = m(B) = 2.7kgAs m(B) moves down, determine the magnitude of the acceleration of m(A)and m(B), given theta = 32 degree.What smallest value of u_k will keep the system from accelerating?u_k is... 顯示更多 Suppose the coefficient of kinetic friction between m(A) and the plane in the figure is u_k = 0.10, and that m(A) = m(B) = 2.7kg As m(B) moves down, determine the magnitude of the acceleration of m(A)and m(B), given theta = 32 degree. What smallest value of u_k will keep the system from accelerating? u_k is kinetic friction http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS4_8/PS4_8.htm the picture is problem 62

最佳解答:

• Approximation and Errors(急急急!)
• 暑期作業(數學問題)係小六升中一嫁
• easy 中文+maths

• Approximation and Errors(急急急!)
• 暑期作業(數學問題)係小六升中一嫁
• easy 中文+maths

 

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The acceleration of the masses can be found from the method given in the link, acceleration a = g[1-sin(theta) - u.cos(theta)]/2 -------- (1) where g is the acceleration due to gravity(=9.8 m/s2) u is the coefficient of kinetic friction(=0.1) "theta" is the slope angle (=32 degrees) hence, a = (9.8)[1-sin(32)- 0.1.cos(32)]/2 m/s2 = 1.89 m/s2 If the acceleration a = 0 m/s2, then equation (1) becomes, 0 = g[1-sin(theta) - u.cos(theta)]/2 i.e. 1-sin(theta) - u.cos(theta) = 0 u = [1-sin(theta)]/cos(theta) = (1-sin(32))/cos(32) = 0.55

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