(a) the acceleration of the two blocks 1 m/s 2 (b) the tension in the string 2 N
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Question
(a) the acceleration of the two blocks1 m/s2
(b) the tension in the string
2 N Coefficients of Frictiona µs µk Steel on steel 0.74 0.57 Aluminum on steel 0.61 0.47 Copper on steel 0.53 0.36 Rubber on concrete 1.0 0.8 Wood on wood 0.25-0.5 0.2 Glass on glass 0.94 0.4 Waxed wood on wet snow 0.14 0.1 Waxed wood on dry snow - 0.04 Metal on metal (lubricated) 0.15 0.06 Ice on ice 0.1 0.03 Teflon on Teflon 0.04 0.04 Synovial joints in humans 0.01 0.003 a Allvalues are approximate.
Coefficients of Frictiona µs µk Steel on steel 0.74 0.57 Aluminum on steel 0.61 0.47 Copper on steel 0.53 0.36 Rubber on concrete 1.0 0.8 Wood on wood 0.25-0.5 0.2 Glass on glass 0.94 0.4 Waxed wood on wet snow 0.14 0.1 Waxed wood on dry snow - 0.04 Metal on metal (lubricated) 0.15 0.06 Ice on ice 0.1 0.03 Teflon on Teflon 0.04 0.04 Synovial joints in humans 0.01 0.003 a Allvalues are approximate. Making use of the table below, determine the following. (a) the acceleration of the two blocks 1 m/s2 (b) the tension in the string 2 N s k Steel on steel 0.74 0.57 Aluminum on steel 0.61 0.47 Copper on steel 0.53 0.36 Rubber on concrete 1.0 0.8 Wood on wood 0.25-0.5 0.2 Glass on glass 0.94 0.4 Waxed wood on wet snow 0.14 0.1 Waxed wood on dry snow - 0.04 Metal on metal (lubricated) 0.15 0.06 Ice on ice 0.1 0.03 Teflon on Teflon 0.04 0.04 Synovial joints in humans 0.01 0.003 a Allvalues are approximate. Coefficients of Frictiona A 1.00 kg aluminum block and a6.00 kg copper block are connected by a light string over a frictionless pulley. The two blocks areallowed to move on a fixed steel block wedge (of angle ? = 45.0 degree) as shown in the figure below.
Explanation / Answer
For aluminum block T - f1 = m1 a T - 0.47 (m1 g) = m1a .......1 For copper block T = m2 g sin - f2 - m2 a = m2 g sin -(0.36) (m2 g cos) - m2a .........2 From eq 1 and 2 we have a = [m2* gsin - 0.36 * m2 * gcos - 0.47 * m1 * g ] / [ m1 + m2 ] Substitute values m1 =1.00 kg m2 =6.00 kg g = 9.8 m/s2 = 45o Solve for acceleration 'a' . b) Now substitute 'a' in eq.1 we get tension T.
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