Question

two applied forces are exerted on a box that slides along a flat surface with negligible friction. the force on the right is at an angle \\( \theta \\) to the horizontal, as shown. the box is initially moving to the right at a constant velocity. the angle \\( \theta \\) is then slowly decreased, but the magnitudes of the applied forces remain constant. which of the following correctly describes the motion of the box as the angle is decreased?
a the velocity of the box remains constant because the sum of all the forces on the box remains constant.
b the velocity of the box remains constant because the horizontal components of the forces do not change.
c the velocity of the box changes because the sum of the vertical components of the forces no longer equals zero.
d the velocity of the box changes because the sum of the horizontal components of the forces no longer equals zero.

Explanation:

Step1: Analyze initial state

Initially, the box moves at constant velocity, so the sum of horizontal forces is zero. Let the left - ward force be \(F_1\) and the right - ward force be \(F_2\). Initially, \(F_1 = F_2\cos\theta\).

Step2: Consider \(\theta\) decreasing

As \(\theta\) decreases, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) (using the definition of cosine in a right - triangle for force components). Since \(F_2\) is constant and \(\theta\) decreases, \(F_{2x}=F_2\cos\theta\) increases (\(\cos\theta\) increases as \(\theta\) decreases from an acute angle). The left - ward force \(F_1\) remains constant. So the sum of horizontal forces \(F_{net,x}=F_2\cos\theta - F_1\) is no longer zero. According to Newton's second law \(F = ma\), if \(F_{net,x}
eq0\), \(a
eq0\) and the velocity changes.

Answer:

D. The velocity of the box changes because the sum of the horizontal components of the forces no longer equals zero.