Find Weight of Object Pulled by Rope w/ FT=30N & Acceleration=1.4m/s2

[ThomasMagnus]

An object is pulled up by a rope from an elevator traveling upward with an acceleration of 1.4 m/s². The tension in the rope is 30N. What is the weight of the object.
A) 2.6kg
B) 3.6kg
C)4.6kg
D)5.6kg

My attempt:
F_T=30N
a_system=1.4m/s²
Weight of object=mg

The system is accelerating up, so let 'up' be positive.
[tex]\Sigma[/tex]F=ma

Sum of the forces acting upwards minus sum of the forces acting downward:
30N+(-mg)=ma
30N=m(a+g)
30/(9.8+1.4)=m
m=2.6kg

For some reason I am not getting the correct answer. The book says that it is 3.6kg. What am I doing wrong?

This is the method the book uses:
"Since the forces act in opposite directions, they are related as: mg=ma+T"

 

[PhanthomJay]

looks like a double book error. 1.) Mass and weight are not the same. 2.) The acceleration and movement are not necessarily in the same direction, and this is not stated. Assuming that the acceleration is upward, what is the object's weight?

 

[ThomasMagnus]

If the acceleration is upward and they are talking about mass, did I do it correctly?

 

[PhanthomJay]

yes.

 

[rwisz]

Therefore,
m= (T/a) - g
m= (30N/1.4m/s^2) - 9.8m/s^2
m= 21.43 - 9.8
m= 11.63kg

It seems that the book has used the equation F=ma, where F is the net force, to solve for the weight (mg) of the object. In your attempt, you have used the equation \SigmaF=ma, where \SigmaF is the sum of all forces, and have solved for the mass (m) of the object. Both approaches are valid, but the book's method may be simpler and more straightforward. It is important to note that the weight of an object is equal to the force of gravity acting on it, which is given by the equation F=mg. Therefore, the weight of the object is 30N, since that is the force of gravity acting on it. The book's method takes into account the tension in the rope, which is an external force acting on the object, while your method only considers the net force. Both methods are correct, but the book's method may be more appropriate in this scenario.