Question

question 14 (1 point)
a container of water is lifted vertically 3.0 m then returned to its original position. if the total weight is 30 n, how much work was done?
45 j
90 j
180 j
no work was done.

Explanation:

Step1: Recall the work formula

The work done by a force \( F \) over a displacement \( d \) in the direction of the force is given by \( W = Fd\cos\theta \), where \( \theta \) is the angle between the force and displacement. When lifting the container up, the force (equal to weight \( F = 30\ N \)) and displacement \( d_1 = 3.0\ m \) are in the same direction (\( \theta = 0^\circ \), \( \cos\theta = 1 \)), so work done in lifting is \( W_1 = Fd_1\cos0^\circ=30\ N\times3.0\ m\times1 = 90\ J \).

Step2: Work done in lowering

When lowering the container back to original position, the force (weight, acting downward) and displacement (upward) are in opposite directions (\( \theta = 180^\circ \), \( \cos\theta=- 1 \)). Displacement \( d_2 = 3.0\ m \), so work done in lowering is \( W_2 = Fd_2\cos180^\circ=30\ N\times3.0\ m\times(- 1)= - 90\ J \).

Step3: Total work done

Total work \( W = W_1+W_2=90\ J+( - 90\ J)=0\ J \). Alternatively, since the net displacement is \( 0\ m \) (returns to original position), using \( W = Fd\cos\theta \) with net displacement \( d = 0 \), we get \( W = 30\ N\times0\ m\times\cos\theta = 0\ J \). So no net work is done.

Answer:

No work was done.