Question

1. a 30.4 newton force is used to slide a 40.0 newton crate a distance of 6.00 meters at constant speed along an incline to a vertical height of 3.00 meters.

a. determine the total work done by the 30.4 - newton force in sliding the crate along the incline.

b. calculate the total increase in the gravitational potential energy of the crate after it has slid 6.00 meters along the incline.

Explanation:

Part a

Step1: Recall work formula

The formula for work done by a force is \( W = F \cdot d \cdot \cos\theta \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement. Here, the force is parallel to the incline, so \( \theta = 0^\circ \) and \( \cos(0^\circ)=1 \).

Step2: Substitute values

Given \( F = 30.4\,\text{N} \), \( d = 6.00\,\text{m} \), and \( \cos(0^\circ) = 1 \). So \( W = 30.4\,\text{N} \times 6.00\,\text{m} \times 1 \).

Step3: Calculate work

\( W = 30.4\times6.00 = 182.4\,\text{J} \) (rounded appropriately, maybe to three significant figures as given in the problem).

Step1: Recall gravitational potential energy formula

The formula for gravitational potential energy change is \( \Delta U = mgh \), but since weight \( W = mg \), we can also write \( \Delta U = W \cdot h \), where \( W \) is the weight of the crate and \( h \) is the vertical height gained.

Step2: Substitute values

Given \( W = 40.0\,\text{N} \), \( h = 3.00\,\text{m} \). So \( \Delta U = 40.0\,\text{N} \times 3.00\,\text{m} \).

Step3: Calculate potential energy change

\( \Delta U = 40.0\times3.00 = 120.0\,\text{J} \)

Answer:

The total work done is \( \boldsymbol{182\,\text{J}} \) (or \( 182.4\,\text{J} \))

Part b