Asked by Joe
An 11 kg object moves down a 25 degree incline at a constant speed. what is the magnitude of the coefficient of friction?
I have m=11.0kg
incline = 25 deg
but i feel like I am missing one more number in order to calculate
I have m=11.0kg
incline = 25 deg
but i feel like I am missing one more number in order to calculate
Answers
Answered by
Ava
Fg = mg
Fg = (11.0)(9.81)
Fg = 107.91 N
Sin x Fg parallel
------ = -------------
1 Fg
Sin 25.0 Fg parallel
--------- = -------------
1 107.91
Cross multiply
(107.91)sin(25) = 45.6 N
≈ 46 N
*(≈ <-- is an approximation symbol)
Fg = (11.0)(9.81)
Fg = 107.91 N
Sin x Fg parallel
------ = -------------
1 Fg
Sin 25.0 Fg parallel
--------- = -------------
1 107.91
Cross multiply
(107.91)sin(25) = 45.6 N
≈ 46 N
*(≈ <-- is an approximation symbol)
Answered by
Ava
Fg = mg
Fg = (11.0)(9.81)
Fg = 107.91 N
Sin x/1 = Fg parallel/Fg
Sin 25.0/1 = Fg parallel/107.91
Cross multiply
(107.91)sin(25) = 45.6 N
≈ 46 N
*(≈ <-- is an approximation symbol)
Fg = (11.0)(9.81)
Fg = 107.91 N
Sin x/1 = Fg parallel/Fg
Sin 25.0/1 = Fg parallel/107.91
Cross multiply
(107.91)sin(25) = 45.6 N
≈ 46 N
*(≈ <-- is an approximation symbol)
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