91Ó°ÊÓ

Work done by a force is calculated using the formula: \ \( W = F \times d \times \text{cos}(\theta) \) \ Where: \ \( F \) is the force applied (7.68 N), \ \( d \) is the distance moved (4.06 m), \ \( \theta \) is the angle of the force above the horizontal (15.0 degrees). \ First, convert the angle to radians if necessary, and then apply the values to the formula: \ \( W = 7.68 \times 4.06 \times \text{cos}(15.0^{\text{circ}}) \) \ Calculate \( W \).

Since the block is moving at constant speed, the work done by the rope's force is equal to the work done against the friction. This means the entire work done by the rope's force is converted into thermal energy. So, the increase in thermal energy \( \text{ΔE}_{\text{th}} \) is equal to the work calculated in Step 1.

The force of friction \( F_{\text{fric}} \) can be calculated from the horizontal component of the rope's force because the block moves at a constant speed. The horizontal component \( F_{\text{horizontal}} \) is given by: \ \( F_{\text{horizontal}} = F \times \text{cos}(\theta) \) \ Set this equal to the friction force: \ \( F_{\text{fric}} = 7.68 \times \text{cos}(15.0^{\text{circ}}) \) \ Next, recall that friction force \( F_{\text{fric}} \) is also given by: \ \( F_{\text{fric}} = \text{μ}_{\text{k}} \times N \) \ where \( \text{μ}_{\text{k}} \) is the coefficient of kinetic friction, and \( N \) is the normal force. The normal force \( N \) can be calculated using Newton's second law in the vertical direction: \ \( N = mg - F \times \text{sin}(\theta) \) \ Plug in the values: \ \( N = 3.57 \times 9.8 - 7.68 \times \text{sin}(15.0^{\text{circ}}) \) \ Finally, solve for \( \text{μ}_{\text{k}} \) by rearranging \( F_{\text{fric}} = \text{μ}_{\text{k}} \times N \) to \( \text{μ}_{\text{k}} = \frac{F_{\text{fric}}}{N} \).