Asked by ~christina~
                2 crates shown in the figure below are on a horizontal frictionless surface. The woman is wearing golf shoes for traction as she applies a horizontal force of <b>F= 50.0 N</b> to the <b>6.00kg</b> crate. The ropes are of negligible mass. Smaller box weighs <b>4.00kg</b>
a) draw a free body diagram for each of the crates and the woman and for each force, indicate what body exerts that force.
b) What is the acceleration of the 6.00kg crate?
c) what is the tension T in the rope connecting the 2 crates.
drawing shows
o-T--0--->person
(the first is a small box and in between the bigger box is T for tension and the arrow is the force direction without a angle..basically directly horizontal to the x axis if I can describe it like that.
~my thinking..
a.) not sure about the woman's free body diagram
_______________________________________
b.)
F= 50N
a=?
m= smaller box + larger box = 4.00kg + 6.00kg= 10.00kg= 10,000g
F= ma
50N= 10,000g*a
a= 5e-3N/g
Is this fine?
_______________________________
c.)
How do I find the tension in the rope between the 2 boxes?
~Thank you~
            
        a) draw a free body diagram for each of the crates and the woman and for each force, indicate what body exerts that force.
b) What is the acceleration of the 6.00kg crate?
c) what is the tension T in the rope connecting the 2 crates.
drawing shows
o-T--0--->person
(the first is a small box and in between the bigger box is T for tension and the arrow is the force direction without a angle..basically directly horizontal to the x axis if I can describe it like that.
~my thinking..
a.) not sure about the woman's free body diagram
_______________________________________
b.)
F= 50N
a=?
m= smaller box + larger box = 4.00kg + 6.00kg= 10.00kg= 10,000g
F= ma
50N= 10,000g*a
a= 5e-3N/g
Is this fine?
_______________________________
c.)
How do I find the tension in the rope between the 2 boxes?
~Thank you~
Answers
                    Answered by
            bobpursley
            
    the woman has a force pulling her back (ie, the boxes are pulling back with 50N, and the shoes on the floor are pulling forward 50N.)
When doing F=ma, M is in kg
a=50/10= 5m/s^2
The acceleration of the smaller box is the same, 5m/s^2, so T= ma=4kg*5m/s^2=20N
    
When doing F=ma, M is in kg
a=50/10= 5m/s^2
The acceleration of the smaller box is the same, 5m/s^2, so T= ma=4kg*5m/s^2=20N
                    Answered by
            ~christina~
            
    Oh okay Thanks Bob
    
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