Question

1. kevin works as a janitor, and he is pushing a fully - loaded recycling bin that weighs 557 n across the floor. he applies a force of 410 n at an angle of 15° below the horizontal. the coefficient of kinetic friction between the recycling bin and the floor is 0.46. how long does it take him to move the bin 6.5 meters, assuming the bin starts from rest?

2.85 s
1.98 s
3.77 s
1.02 s

Explanation:

Step1: Find the mass of the bin

We know that weight \(W = mg\). Given \(W=557\ N\) and \(g = 9.8\ m/s^{2}\), then \(m=\frac{W}{g}=\frac{557}{9.8}\ kg\approx56.84\ kg\)

Step2: Analyze the vertical forces

The normal force \(N\) has contributions from the weight of the bin and the vertical component of the applied force. The applied force \(F = 410\ N\) at an angle \(\theta=15^{\circ}\) below the horizontal.
The vertical component of the applied force \(F_y=F\sin\theta\). So \(N=W + F\sin\theta\)
\(N = 557+410\sin15^{\circ}\)
\(\sin15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259\)
\(N=557 + 410\times0.259=557+106.19=663.19\ N\)

Step3: Calculate the frictional force

The frictional force \(f=\mu_kN\), where \(\mu_k = 0.46\)
\(f=0.46\times663.19\ N\approx305.07\ N\)

Step4: Analyze the horizontal forces and find the acceleration

The horizontal component of the applied force \(F_x=F\cos\theta\), \(F\cos\theta=410\cos15^{\circ}\)
\(\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966\)
\(F_x = 410\times0.966 = 396.06\ N\)
Using Newton's second law \(F_{net,x}=ma\), \(F_x - f=ma\)
\(a=\frac{F_x - f}{m}=\frac{396.06- 305.07}{56.84}=\frac{90.99}{56.84}\approx1.6\ m/s^{2}\)

Step5: Use the kinematic equation \(x = v_0t+\frac{1}{2}at^{2}\)

Since \(v_0 = 0\) (starts from rest) and \(x = 6.5\ m\), the equation becomes \(x=\frac{1}{2}at^{2}\)
\(t=\sqrt{\frac{2x}{a}}\)
Substitute \(x = 6.5\ m\) and \(a\approx1.6\ m/s^{2}\)
\(t=\sqrt{\frac{2\times6.5}{1.6}}=\sqrt{\frac{13}{1.6}}=\sqrt{8.125}\approx2.85\ s\)

Answer:

2.85 s