Question

question 11 (1 point)
a 30-n box is pulled 6.0 m up along a 37° inclined plane. what is the work done by
the weight (gravitational force) of the box?

• 11 j
• 1.1 × 10^2j
• 1.4 × 10^2 j
• 1.8 × 10^2 j

Explanation:

Step1: Recall the work formula

The work done by a force is given by \( W = Fd\cos\theta \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the displacement.

Step2: Identify the force and angle

The weight of the box is \( F = 30\,\text{N} \) (acting vertically downward). The displacement \( d = 6.0\,\text{m} \) is along the inclined plane. The angle between the weight (vertical) and the displacement (along the incline) is \( 90^\circ + 37^\circ = 127^\circ \)? Wait, no. Wait, the component of weight along the incline is \( mg\sin\theta \) (opposing the motion), but actually, the angle between the weight vector (downward) and the displacement vector (up the incline) is \( 90^\circ + 37^\circ \)? Wait, no. Let's think again. The displacement is up the incline, length \( d = 6.0\,\text{m} \). The weight is vertical. The angle between the weight (force) and the displacement (direction) is \( 90^\circ + 37^\circ \)? Wait, no. The angle between the force (weight, downward) and the displacement (up the incline) is \( 180^\circ - 53^\circ \)? Wait, maybe better to find the component of weight along the direction opposite to displacement. The weight has a component along the incline downward: \( F_{\parallel} = mg\sin\theta \), where \( mg = 30\,\text{N} \), \( \theta = 37^\circ \). The displacement is up the incline, so the angle between the force component (down the incline) and displacement (up the incline) is \( 180^\circ \). Wait, no. The work done by weight is the force (weight) times displacement times cosine of the angle between them. The weight is vertical, displacement is along the incline (37 degrees above horizontal). So the angle between weight (vertical) and displacement (37 degrees above horizontal) is \( 90^\circ + 37^\circ = 127^\circ \)? Wait, no. Let's calculate the angle between the two vectors. The displacement vector is along the incline (direction: 37 degrees from horizontal, up). The weight vector is vertical (direction: 90 degrees from horizontal, down). So the angle between them is \( 90^\circ + 37^\circ = 127^\circ \). But \( \cos(127^\circ) = \cos(180^\circ - 53^\circ) = -\cos(53^\circ) \approx -0.6 \). Alternatively, we can think of the component of weight along the displacement direction. The displacement is up the incline, the weight has a component down the incline: \( F_{\parallel} = 30\,\text{N} \times \sin(37^\circ) \). Then the work done by weight is \( W = -F_{\parallel} \times d \) (negative because force and displacement are opposite). Wait, \( \sin(37^\circ) \approx 0.6 \), so \( F_{\parallel} = 30 \times 0.6 = 18\,\text{N} \) down the incline. Displacement is 6.0 m up the incline, so the angle between force (18 N down) and displacement (6 m up) is 180 degrees. So \( W = Fd\cos\theta = 30\,\text{N} \times 6.0\,\text{m} \times \cos(127^\circ) \). Wait, \( \cos(127^\circ) = \cos(90^\circ + 37^\circ) = -\sin(37^\circ) \approx -0.6 \). So \( W = 30 \times 6.0 \times (-0.6) = 30 \times 6 \times (-0.6) = 180 \times (-0.6) = -108\,\text{J} \approx -1.1 \times 10^2\,\text{J} \).

Step3: Calculate the work

Using \( W = Fd\cos\theta \), where \( F = 30\,\text{N} \), \( d = 6.0\,\text{m} \), \( \theta = 180^\circ - 53^\circ = 127^\circ \) (angle between weight and displacement). \( \cos(127^\circ) \approx -0.6 \). So \( W = 30 \times 6.0 \times (-0.6) = -108\,\text{J} \approx -1.1 \times 10^2\,\text{J} \).

Answer:

• \( 1.1 \times 10^2\,\text{J} \) (with the negative sign, so the option is - \( 1.1 \times 10^2\,\text{J} \))