A body of mass 10kg rests on a surface travelling downwards with an accelaration of 3m/s^2. Determine the apparent weight.
Please show calculations
Please show calculations
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You don't say upon what planet you are performing this experiment.
If you are on earth where g = 9.8 m/s^2
you may either calculate the relative acceleration ( g-a = 6.8 m/s^2)
and use this F = 6.8 * 10 = 68 N
Or you may look at the forces.
The NET force is making it accelerate.
Hence the net force causes an acceleration of 3 m/s
The net force is the force of gravity - force of floor on object
hence mg - apparent Weight = ma
9.8 m - Weight = 3 m
Weight = 9.8 m - 3m = 6.8 m = 68 N Just as I obtained above.
If you are on earth where g = 9.8 m/s^2
you may either calculate the relative acceleration ( g-a = 6.8 m/s^2)
and use this F = 6.8 * 10 = 68 N
Or you may look at the forces.
The NET force is making it accelerate.
Hence the net force causes an acceleration of 3 m/s
The net force is the force of gravity - force of floor on object
hence mg - apparent Weight = ma
9.8 m - Weight = 3 m
Weight = 9.8 m - 3m = 6.8 m = 68 N Just as I obtained above.
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apparent weight R = m ( g - a) = 10 ( 9.8 - 3 ) = 63 N.
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what`s `apparent` weight?never heard of it...
we have weight,who splits in 2 and friction in this case...
something else?maybe `imaginary` weight?
we have weight,who splits in 2 and friction in this case...
something else?maybe `imaginary` weight?