Question

a 10 kg box is pulled along a horizontal surface with a force of 25 n at an angle of 30 degrees above the horizontal. if the coefficient of kinetic friction is 0.2, what is the acceleration of the box? draw a free - body diagram.

Explanation:

Step1: Analyze Forces in Vertical Direction

The vertical forces on the box are the normal force \( N \), the vertical component of the pulling force \( F_y = F\sin\theta \), and the weight \( mg \). In vertical equilibrium (no acceleration vertically), \( N + F\sin\theta = mg \). So, \( N = mg - F\sin\theta \). Substituting values: \( m = 10\,\text{kg} \), \( g = 9.8\,\text{m/s}^2 \), \( F = 25\,\text{N} \), \( \theta = 30^\circ \). \( N = 10\times9.8 - 25\times\sin30^\circ = 98 - 12.5 = 85.5\,\text{N} \).

Step2: Calculate Frictional Force

The kinetic frictional force \( f_k = \mu_k N \), where \( \mu_k = 0.2 \). So, \( f_k = 0.2\times85.5 = 17.1\,\text{N} \).

Step3: Analyze Forces in Horizontal Direction

The horizontal forces are the horizontal component of the pulling force \( F_x = F\cos\theta \) and the frictional force \( f_k \). Using Newton's second law \( F_{net,x} = ma \), so \( F\cos\theta - f_k = ma \).

Step4: Solve for Acceleration

Substitute \( F\cos\theta = 25\times\cos30^\circ \approx 25\times0.866 = 21.65\,\text{N} \), \( f_k = 17.1\,\text{N} \), \( m = 10\,\text{kg} \) into the equation: \( 21.65 - 17.1 = 10a \). \( 4.55 = 10a \), so \( a = \frac{4.55}{10} = 0.455\,\text{m/s}^2 \approx 0.46\,\text{m/s}^2 \).

Free - Body Diagram:

• Forces on the box:
• Gravitational force (\( mg \)): Downward, magnitude \( 10\times9.8 = 98\,\text{N} \).
• Normal force (\( N \)): Upward, magnitude \( 85.5\,\text{N} \) (calculated above).
• Pulling force (\( F \)): At an angle of \( 30^\circ \) above horizontal, magnitude \( 25\,\text{N} \). It can be resolved into horizontal (\( F\cos30^\circ \)) and vertical (\( F\sin30^\circ \)) components.
• Kinetic frictional force (\( f_k \)): Horizontal, opposite to the direction of motion, magnitude \( 17.1\,\text{N} \).

Answer:

The acceleration of the box is approximately \( \boldsymbol{0.46\,\text{m/s}^2} \) (and the free - body diagram is as described above).