= 24 ore1260mittVa = 15V.V = 30v
1 answer:
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Answer:
a) 2.53 * 10^-2 m/s
b) -4.78 * 10^-2 m/s
c) 1.21 * 10^-1 m/s
Explanation:
Given data :
Mass of block = 10 kg
Measuring 250mm on each side
a) calculate the speed when a force of 75N is applied to pull block upwards
F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis ) ----- ( 1 )
f ( friction force )= ( 16400v * 6.25 *10^-2) = 1025 v
F ( force applied ) = 75
W ( weight of block ) = 10 * 9.81 = 98.1 N
∅ = 30°
input values into equation 1
V = = 2.53 * 10^-2 m/s
b) Speed when no force is applied on the block
F = f + W sin∅
F = 0
f = 1025 V
W = 98.1 N
∅ = 30°
hence V = = - 4.78 * 10^-2 m/s
c) when a force is applied to push block down the incline
F = f + W sin∅ ----- ( 3 )
F = 75 N
f = 1025 V
W = 98.1 N
∅ = 30°
input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction
75 = 1025 V - 98.1 ( sin 30° )
V = = 1.21 * 10^-1 m/s