Question

shown in the figure. frictional forces between the block and the surface are negligible. at time ( t = t_1 ), the angle ( \theta ) between the force ( f ) and the horizontal starts to increase without exceeding ( 90^{circ} ), while the magnitude of the force remains constant. how does the kinetic energy of the object change as the angle increases?

a the kinetic energy remains the same.
b the kinetic energy increases.
c the kinetic energy decreases.
d the kinetic energy increases when ( \theta < 45^{circ} ) and decreases when ( \theta > 45^{circ} ).

incorrect
c
incorrect. considering the work - energy theorem, even though rate at which work is being done on the block decreases, there is always positive work being done on the block by the force ( f ) because there is always a component of the force exerted in the same direction as the motion of the block.

Explanation:

Step1: Analyze the horizontal component of force

The horizontal component of force \(F\) is \(F_{x}=F\cos\theta\). As \(\theta\) increases from \(0\) to \(90^{\circ}\), \(\cos\theta\) decreases (\(\cos\theta\) is a decreasing function in the interval \([0,90^{\circ}]\)). But \(F_{x}=F\cos\theta> 0\) as long as \(\theta<90^{\circ}\).

Step2: Apply the work - energy theorem

The work - energy theorem states that \(W=\Delta K\), where \(W\) is the work done on the object and \(K\) is the kinetic energy. The work done by the force \(F\) is \(W = F_{x}d=Fd\cos\theta\) (where \(d\) is the displacement in the horizontal direction). Since \(F>0\), \(d > 0\) (the block is moving) and \(\cos\theta>0\) for \(0\leq\theta<90^{\circ}\), the work done \(W>0\).

Answer:

B. The kinetic energy increases.