Question

question 46
two forces act on a 4.5 - kg block on a frictionless surface as shown. what is the magnitude of the horizontal acceleration of the block?

8.9 m/s²
3.2 m/s²
1.8 m/s²
1.2 m/s²
0.82 m/s²

Explanation:

Step1: Calculate the horizontal component of the \(5.9N\) force

The horizontal component of a force \(F\) at an angle \(\theta\) is \(F\cos\theta\). For \(F = 5.9N\) and \(\theta=43^{\circ}\), the horizontal component is \(F_{x1}=5.9\cos(43^{\circ})\). Using \(\cos(43^{\circ})\approx0.731\), we get \(F_{x1}=5.9\times0.731 = 4.3129N\).

Step2: Find the net horizontal force

The other horizontal force is \(F_{x2} = 3.7N\). The net horizontal force \(F_{net}=F_{x1}+F_{x2}\). Substituting the values, \(F_{net}=4.3129 + 3.7=8.0129N\).

Step3: Use Newton's second law \(F = ma\) to find acceleration

Given \(m = 4.5kg\) and \(F_{net}=8.0129N\), and \(a=\frac{F_{net}}{m}\). So \(a=\frac{8.0129}{4.5}\approx1.8m/s^{2}\).

Answer:

\(1.8m/s^{2}\) (the third option)