QuestionJuly 7, 2025

A tension force of 155 N inclined at 15.0^circ above the horizontal is used to pull a 35.0 kg storage crate a distance of 4.10 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b)the coefficient of kinetic friction between the crate and surface. HINT (a) the work done by the tension force (in J) square J (b) the coefficient of kinetic friction between the crate and surface square

A tension force of 155 N inclined at 15.0^circ above the horizontal is used to pull a 35.0 kg storage crate a distance of 4.10 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b)the coefficient of kinetic friction between the crate and surface. HINT (a) the work done by the tension force (in J) square J (b) the coefficient of kinetic friction between the crate and surface square
A tension force of 155 N inclined at 15.0^circ above the horizontal is used to pull a 35.0 kg storage crate a distance of 4.10 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b)the coefficient of kinetic friction between the crate and surface. HINT (a) the work done by the tension force (in J) square J (b) the coefficient of kinetic friction between the crate and surface square

Solution
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Answer

(a) 613 J ### (b) 0.252 Explanation 1. Calculate the work done by the tension force Work is calculated using **W = F \cdot d \cdot \cos(\theta)**. Here, F = 155 \, \text{N}, d = 4.10 \, \text{m}, and \theta = 15.0^\circ. Thus, W = 155 \cdot 4.10 \cdot \cos(15.0^\circ). 2. Calculate the horizontal component of the tension force The horizontal component is **F_{\text{horizontal}} = F \cdot \cos(\theta)**. So, F_{\text{horizontal}} = 155 \cdot \cos(15.0^\circ). 3. Calculate the normal force Normal force N = mg - F \cdot \sin(\theta), where m = 35.0 \, \text{kg} and g = 9.81 \, \text{m/s}^2. Thus, N = 35.0 \cdot 9.81 - 155 \cdot \sin(15.0^\circ). 4. Calculate the frictional force Since the crate moves at constant speed, frictional force f_k = F_{\text{horizontal}}. 5. Calculate the coefficient of kinetic friction Coefficient of kinetic friction \mu_k = \frac{f_k}{N}.

Explanation

1. Calculate the work done by the tension force<br /> Work is calculated using **$W = F \cdot d \cdot \cos(\theta)$**. Here, $F = 155 \, \text{N}$, $d = 4.10 \, \text{m}$, and $\theta = 15.0^\circ$. Thus, $W = 155 \cdot 4.10 \cdot \cos(15.0^\circ)$.<br /><br />2. Calculate the horizontal component of the tension force<br /> The horizontal component is **$F_{\text{horizontal}} = F \cdot \cos(\theta)$**. So, $F_{\text{horizontal}} = 155 \cdot \cos(15.0^\circ)$.<br /><br />3. Calculate the normal force<br /> Normal force $N = mg - F \cdot \sin(\theta)$, where $m = 35.0 \, \text{kg}$ and $g = 9.81 \, \text{m/s}^2$. Thus, $N = 35.0 \cdot 9.81 - 155 \cdot \sin(15.0^\circ)$.<br /><br />4. Calculate the frictional force<br /> Since the crate moves at constant speed, frictional force $f_k = F_{\text{horizontal}}$. <br /><br />5. Calculate the coefficient of kinetic friction<br /> Coefficient of kinetic friction $\mu_k = \frac{f_k}{N}$.
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