Questions: A box initially at rest, is pushed with a force of 16.3 N at a 41.0° angle over a distance of 9.32 m. A frictional force of 6.23 N is exerted back onto the box. What is the total amount of work done? w=[?] J

Solution Steps

The work done by a force is given by the formula:

\[ W = F \cdot d \cdot \cos(\theta) \]

where:

• \( F = 16.3 \, \text{N} \) is the magnitude of the applied force,
• \( d = 9.32 \, \text{m} \) is the distance over which the force is applied,
• \( \theta = 41.0^\circ \) is the angle between the force and the direction of motion.

Substituting the given values:

\[ W_{\text{applied}} = 16.3 \cdot 9.32 \cdot \cos(41.0^\circ) \]

Calculating the cosine of \(41.0^\circ\):

\[ \cos(41.0^\circ) \approx 0.7547 \]

Now, calculate the work done by the applied force:

\[ W_{\text{applied}} = 16.3 \cdot 9.32 \cdot 0.7547 \approx 114.7 \, \text{J} \]

The work done by the frictional force is calculated using:

\[ W_{\text{friction}} = F_{\text{friction}} \cdot d \cdot \cos(180^\circ) \]

where:

• \( F_{\text{friction}} = 6.23 \, \text{N} \) is the magnitude of the frictional force,
• \( \cos(180^\circ) = -1 \) because the frictional force opposes the direction of motion.

Substituting the given values:

\[ W_{\text{friction}} = 6.23 \cdot 9.32 \cdot (-1) = -58.0 \, \text{J} \]

The total work done on the box is the sum of the work done by the applied force and the work done by the frictional force:

\[ W_{\text{total}} = W_{\text{applied}} + W_{\text{friction}} \]

Substituting the calculated values:

\[ W_{\text{total}} = 114.7 + (-58.0) = 56.7 \, \text{J} \]

Final Answer