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 vertical forces

The vertical forces are the normal force \(N\), the vertical component of the applied force \(F\sin\theta\), and the weight \(mg\). Using Newton's second law in the vertical direction (\(a_y = 0\)), we have \(N+F\sin\theta=mg\). So, \(N = mg - F\sin\theta\). Substituting \(m = 10\space kg\), \(g=9.8\space m/s^{2}\), \(F = 25\space N\), and \(\theta = 30^{\circ}\), we get \(N=10\times9.8 - 25\times\sin30^{\circ}=98 - 12.5=85.5\space N\).

Step2: Calculate frictional force

The kinetic frictional force \(f_k=\mu_kN\). Given \(\mu_k = 0.2\) and \(N = 85.5\space N\), then \(f_k=0.2\times85.5 = 17.1\space N\).

Step3: Analyze horizontal forces

Using Newton's second law in the horizontal direction (\(F_{net,x}=ma_x\)), the net force \(F_{net,x}=F\cos\theta - f_k\). Substituting \(F = 25\space N\), \(\theta = 30^{\circ}\), and \(f_k = 17.1\space N\), we have \(F_{net,x}=25\cos30^{\circ}-17.1\). Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\), \(F_{net,x}=25\times0.866-17.1=21.65 - 17.1 = 4.55\space N\).

Step4: Find acceleration

From \(F_{net,x}=ma_x\), we can solve for \(a_x\). Given \(m = 10\space kg\) and \(F_{net,x}=4.55\space N\), then \(a_x=\frac{F_{net,x}}{m}=\frac{4.55}{10}=0.455\space m/s^{2}\).

Answer:

The acceleration of the box is \(0.455\space m/s^{2}\).