What does a 124 rock weigh if it is accelerating upward at 19.0 ?
do you have units for original weight of rock and acceleration?
sorry yes, 124 N and 19.0 m/s^2
a a newton is defined as: a kilogram meter per second squared \[\text N=\frac{\text{kg}\cdot\text m}{\text s^2}\]
I disagree. Think of riding in an elevator accelerating upwards. Your legs have to take up the extra push from the elevator floor as the elevator travels (i.e., your weight should be greater not less). If a rock weighs 124N while at rest on the surface of the Earth, it's mass is 124/9.8=12.7kg. Now, it's weight as measured while accelerating upward will be \[W=12.7*28.8=364.4N\]
But, if I misunderstood your question and you are asking instead "what is the weight of the rock while at rest on the Earth's surface if it weights 124N while accelerating 19m/s^2 upward, away from the surface of the Earth", then we have for the mass of the rock, \[124N/28.8=4.3 kg\]This gives a weight of:\[W=4.3*9.8=42.2N\]while at rest at the surface of the Earth.
basically, 124N means mass is 124/9.8 kg, so upward force (net on the body) is 124/9.8 *19, there is no meaning of it's weight changing when it's accelerated... (If you take the frame of reference as the rock, then ofcourse the rock will feel heavier)... In a lift, the net force downwards increases(wrt to lift) hence u feel heavier....
I take an operational definition of weight in which weight is defined to be the value read off a spring (or suitable) balance. In an elevator accelerating upwards, the spring will have a longer extension than when the elevator is at rest (or moving uniformly), hence the weight in the accelerated case is said to be greater than in the non-accelerated case.
the \(effective\) weight of the rock will be 364.4N
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