A box of mass m is pulled with a constant acceleration a along a horizontal frictionless floor by a wire that makes an angle of 15° above the horizontal. If T is the tension in this wire, then _____________.

A) T = ma.
B) T > ma.
C) T < ma.

- The horizontal component of the tension in the wire, [tex]Tcos \theta[/tex], where T is the magnitude of the tension and [tex]\theta[/tex] the angle that the wire makes with the horizontal

Since this is the only force acting on the box in the horizontal direction, this is also the net force, so it is equal to the product of mass and acceleration (Newton's second law of motion):

[tex]Tcos \theta = ma[/tex]

where

m is the mass of the box

a is the acceleration

We can rewrite the equation as

[tex]T=\frac{ma}{cos \theta}[/tex]

The angle in this problem is [tex]\theta=15^{\circ}[/tex], so

[tex]T=\frac{ma}{cos 15^{\circ}}=\frac{ma}{0.966}=1.035 ma[/tex]

Therefore, the correct option is

B) T > ma.

A box of mass m is pulled with a constant acceleration a along a horizontal frictionless floor by a wire that makes an angle of 15° above the horizontal. If T is the tension in this wire, then _____________. A) T = ma. B) T > ma. C) T < ma.

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A box of mass m is pulled with a constant acceleration a along a horizontal frictionless floor by a wire that makes an angle of 15° above the horizontal. If T is the tension in this wire, then _____________.

A) T = ma.
B) T > ma.
C) T < ma.