Answer:
B) T > ma
Explanation:
To solve the problem, we have to apply Newton's second law, which states that the resultant of the forces acting on the box is equal to the product between the mass of the box (m) and its acceleration (a):
[tex]\sum F = ma[/tex] (1)
We are only interested in the forces acting along the horizontal direction (because there is no acceleration along the vertical axis, since the box is sliding along the floor). There are only two forces acting along the horizontal direction:
- The component of the tension of the wire parallel to the floor, [tex]T cos 15^{\circ}[/tex]
- The force of friction, [tex]F_f[/tex], acting in the opposite direction (against the motion)
So, eq.(1) becomes
[tex]T cos 15^{\circ} - F_f = ma[/tex]
which can be rewritten as
[tex]T = \frac{ma+F_f}{cos 15^{\circ}}[/tex]
We can notice that:
[tex]ma + F_f > ma[/tex] by definition
[tex]cos 15^{\circ} < 1[/tex]
This means that the tension T is greater than the product (ma), so the correct answer is B.
The tension in the wire ,T, is greater than 'ma' by 4% (T > ma).
According to Newton's second law of motion, when a force is applied to an object, the object tends to move in the direction of the applied force.
The motion of the object is due to unbalanced force acting on the object.
The net horizontal force on the object is given as;
[tex]T cos(\theta) - F_k = ma\\\\Tcos(\theta) = ma + F_k[/tex]
where;
T is the tension on the string
[tex]F_k[/tex] is the frictional force between the box and the horizontal surface
When the frictional force is zero (frictionless floor )
Tcos(θ) = ma + 0
Tcos(θ) = ma
Tcos(15) = ma
T(0.966) = ma
[tex]T = \frac{ma}{0.966} \\\\T = 1.04 (ma)[/tex]
Thus, we can conclude that the tension in the wire ,T, is greater than 'ma' by 4% (T > ma).
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