Question

011 10.0 points
the block of mass 4.78893 kg has an acceleration of 4.6 m/s² as shown.
what is the magnitude of f? assume the acceleration due to gravity is 9.8 m/s² and the surface is frictionless.
answer in units of n.

Explanation:

Step1: Analyze the horizontal component of force

According to Newton's second law \(F_{net}=ma\). The horizontal component of the force \(F\) is \(F\cos20^{\circ}\) (since the surface is frictionless, this is the only horizontal force). So \(F\cos20^{\circ}=ma\).

Step2: Solve for \(F\)

We know \(m = 4.78893\space kg\), \(a=4.6\space m/s^{2}\), and \(\cos20^{\circ}\approx0.9397\). Rearranging the formula \(F=\frac{ma}{\cos20^{\circ}}\).
Substitute the values: \(F=\frac{4.78893\times4.6}{0.9397}\).
First calculate \(4.78893\times4.6 = 22.029078\).
Then \(F=\frac{22.029078}{0.9397}\approx23.4\space N\).

Answer:

\(23.4\space N\)