Q: A 5.0 kg block is placed on top of a 10. kg block.(DIAGRAM : ww w.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch05/Images/P5.74.g i f) A horizontal force of 45 N is applied to the 10. kg block and the 5.0 kg block is tied to the wall. The coefficient of friction between the moving surfaces is 0.20.
Determine the tension in the string that ties the 5.0 kg block to the wall
so fnet = =Ft+uma
OR is it Fnet = -Ft+umg
reall im asking since the 5kg block is on top and the 10kg block get a Force applied of 45N would we use ma or mg???
Ft=force of friction on top-forceoffriction bottom
To determine the tension in the string that ties the 5.0 kg block to the wall, we need to consider the forces acting on both blocks.
Let's analyze the forces acting on the 10.0 kg block first. Since there is a horizontal force of 45 N being applied to it, we need to consider the force of friction and the force of gravity acting on it.
The force of friction can be calculated using the formula: force of friction = coefficient of friction × normal force. In this case, the coefficient of friction is 0.20. The normal force is equal to the weight of the block, which is mass × gravity. So, the force of friction can be written as: force of friction = 0.20 × (10.0 kg × 9.8 m/s^2) = 19.6 N.
Since the block is in motion, the frictional force opposes the applied force. Therefore, the net force acting on the 10.0 kg block is: net force = applied force - force of friction = 45 N - 19.6 N = 25.4 N.
Now, let's move on to the 5.0 kg block. Since it is tied to the wall, it does not experience any horizontal force. Therefore, the net force acting on the 5.0 kg block is simply the force of friction. The force of friction on the 5.0 kg block is the same as the force of friction on the 10.0 kg block, which is 19.6 N.
Since the 5.0 kg block is tied to the wall, the tension in the string is equal to the force of friction on the 5.0 kg block, which is 19.6 N.
In summary, the tension in the string that ties the 5.0 kg block to the wall is 19.6 N.