Question

1. carrie pulls a 41.5 kg bin across the floor of her garage using a rope that is attached to the bin. she pulls with a force of 138 n at an angle of 28.0° above horizontal. the coefficient of kinetic friction between the floor and the box is 0.290. what is the acceleration of the box?

0.451 m/s²
0.212 m/s²
0.544 m/s²
0.040 m/s²

Explanation:

Step1: Resolve the pulling force

The horizontal component of the force \(F_x = F\cos\theta\), where \(F = 138\ N\) and \(\theta=28.0^{\circ}\). So \(F_x=138\cos(28.0^{\circ})\approx138\times0.883 = 122\ N\)

Step2: Calculate the normal force

The vertical component of the force \(F_y = F\sin\theta=138\sin(28.0^{\circ})\approx138\times0.469 = 64.7\ N\).
The normal force \(N = mg - F_y\), where \(m = 41.5\ kg\) and \(g = 9.8\ m/s^{2}\). So \(N=41.5\times9.8- 64.7=406.7-64.7 = 342\ N\)

Step3: Calculate the frictional force

The frictional force \(f=\mu_kN\), with \(\mu_k = 0.290\). So \(f = 0.290\times342=99.2\ N\)

Step4: Apply Newton's second law

According to \(F_{net}=ma\), and \(F_{net}=F_x - f\). So \(a=\frac{F_x - f}{m}=\frac{122 - 99.2}{41.5}=\frac{22.8}{41.5}\approx0.549\ m/s^{2}\approx0.544\ m/s^{2}\) (due to rounding differences in intermediate steps)

Answer:

\(0.544\ m/s^{2}\) (corresponding to the third option)