A block with mass my -8.6 kg ts on an incline with an angle 0-26 with respect to

Question

A block with mass my -8.6 kg ts on an incline with an angle 0-26 with respect to the hortzontal. For the first question there ts no friction between the incline and the block. 1) When there ts no friction, what ts the magnitude of the acceleration of the block? m/s2 Submit You currently have O submissions for this question. Only 3 submisston are allowed. You can make 3 more submissions for this question. 2) Now with friction, the acceleration ts measured to be only a -2.88 m/s. What ts the coefficient of kinetic friction between the Incline and the block Submit You currently have O submissions for this question. Only 3 submission are allowed. You can make 3 more submissions for this question. 3)

Explanation / Answer

i) acceleration = gsin26° = 4.296 m/s²

ii) a = 2.88 m/s² = gsin26° - gcos26° = 4.296 - 0.89879(9.8)

=> = (4.296 – 2.88)/(0.89879*9.8)= 0.1608

iii) F = -kx

= 158 x 0.13 = 20.54 N (restoring force)

a = 0 = (mgsin26 – F – mgcos26]

=> [8.6 (9.8sin26) - 20.54 – *8.6*9.8cos26] / 9.2 = 0

=> 16.4 - *8.6*9.8cos26 = 0

=> = 0.2166

iv)

acceleration need to be 0

assuming the rope is detached

force down incline = (m+8.6)gsin26°

force up incline = 0.2166(9.8)cos26(8.6) + m(0.73)9.8cos26 = 16.4 + 6.43m

a = 0 => (m+8.6)gsin26° = 16.4 + 6.43m

=> m(6.43 - 4.296) = 36.95 – 16.4 = 20.55

=> m ~= 9.63 kg

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