Two objects 5 kg and 7 kg are attached to the end of inextensible string which passes over frictionless pulley
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Applying conservation of momentum;
Where m1 = 80.0 kg; v1 =4.10 m/s; m2 = 130.0 kg; v2 = 0; mf = (80+130) kg; vf = ??
Therefore,
80*4.1 + 130*0 = 210*vf
vf = (80*4.1)/210 = 1.56 m/s
Where m1 = 80.0 kg; v1 =4.10 m/s; m2 = 130.0 kg; v2 = 0; mf = (80+130) kg; vf = ??
Therefore,
80*4.1 + 130*0 = 210*vf
vf = (80*4.1)/210 = 1.56 m/s
To solve this problem it is necessary to apply the concepts related to the conservation of Energy. Mathematically the conservation of kinetic energy must be paid in the increase of potential energy or vice versa. This expressed in algebraic terms is equivalent to
Kinetic Energy = Potential Energy
Where
m = Mass
v = Velocity
g = Gravity
h = Height
As the mass is the same then we have to
Rearrange to find v,
Our values are given as
Therefore replacing we have
Hence the velocity at the moon would be 4.99m/s
The only direct affectation is that concerning the Resistance or drag force generated by a fluid - such as air in the ground - that can diminish / sharpen the direct effects of gravity. Disregarding the resistance of the air, as we can see in the equation previously given, there should be no affectation because the speed depends on the gravity and height.