Question

the horizontal surface on which the block of mass 4.3 kg slides is frictionless. the force 60 n acts on the block in a horizontal direction and a force of 180 n acts on the block at an angle as shown below. what is the magnitude of the resulting acceleration of the block? the acceleration of gravity is 9.8 m/s².

Explanation:

Step1: Calculate the horizontal component of the 180 N force

The horizontal component of a force \(F\) at an angle \(\theta\) is \(F_{x}=F\cos\theta\). Here, \(F = 180\space N\) and \(\theta = 60^{\circ}\), so \(F_{x}=180\cos60^{\circ}=180\times0.5 = 90\space N\)

Step2: Find the net horizontal force

The net horizontal force \(F_{net}\) is the sum of the 60 N force and the horizontal component of the 180 N force. So \(F_{net}=60 + 90=150\space N\)

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

Newton's second law is \(a=\frac{F}{m}\). Given \(m = 4.3\space kg\) and \(F_{net}=150\space N\), then \(a=\frac{150}{4.3}\approx34.9\space m/s^{2}\)

Answer:

\(a\approx34.9\space m/s^{2}\)