You push a block of dry ice so that it has an acceleration of 0.79 m/s^{2}. If the net force on the block is 2.0 N, what is the mass of the block?
In this problem, you are given information about the acceleration of, and net force on, a block of dry ice. Acceleration and net force are related through F_{net} = ma, and so this is a one step problem. You do not need to through the full process of Newton’s Second Law because all force information is provided in a single (already added) term.
There is no need for a picture in most definition problems, and this is one of them. You are given net force and acceleration information on one object and asked for mass—the property that mediates how much acceleration you get for a given net force. A picture will not provide any additional insight or organization beyond what is already present in the problem.
In equation form, net force and acceleration are related through F_{net} = ma This is the only relation you need for this problem.
F_{net} = ma
2.0 N = m (0.79 m/s^{2})
2.0 N/(0.79 m/s^{2}) = m
m = 2.5 kg
There is no further calculation required in this problem.
F_{net} is the same as ∑F. Use whichever symbol is preferred by your book and your instructor.
The MKS unit of force is the Newton. 1 N = 1 kg m/s^{2}.
2.0 N/(0.79 m/s^{2}) = [2.0 kg m/s^{2}]/(0.79 m/s^{2}) = 2.5 kg
The MKS unit of force is the Newton. 1 N = 1 kg m/s^{2}.
2.0 N/(0.79 m/s^{2}) = [2.0 kg m/s^{2}]/(0.79 m/s^{2}) = 2.5 kg
In this case, you are told the net force on, and acceleration of, the block. You don’t need to draw a free body diagram in order to find the force because you already know the vector on each side of the Newton’s Second Law equation.
F_{net} = ma
2.0 N = m (0.79 m/s^{2})
m = 2.5 kg
It is fairly common to miss a simple one-step force problem by thinking too hard. When you are either given or asked for a net force and given (or asked for) acceleration, you do not need to go through the full process of a Newton’s Second Law problem.