Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2
)
Answer:
108 N
Explanation:
From the question,
Applying
F' = mgμ................ Equation 1
Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.
From the question,
Given: m = 18 kg, μ = 0.6, g = 10 m/s²
Substitute these values into equation 1
F' = 18×0.6×10
F' = 108 N
In a displacement versus time graph, the slope of the line at any point on the graph indicates the <em>magnitude of velocity</em>.
(It can't indicate velocity completely, because the graph shows nothing about the direction of the motion.)
Answer:
Explanation:
F = ma
4.45g - 2.75g = (4.45 + 2.75)a
a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²
a)
T = 2.75(9.81 + 2.32) = 33.3 N
or
T = 4.45(9.81 - 2.32) = 33.3 N
b) 2.32 m/s² upward for 2.75 kg mass
2.32 m/s² downward for 4.45 kg mass
c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m
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