Question

The block of mass 3.37548 kg has an acceleration of 3.2 m/s2as shown.3.37548 kgF30◦aWhat is the magnitude of F? Assume theacceleration due to gravity is 9.8 m/s2and thesurface is frictionless.Answer in units of N.

Answer 1

The magnitude of F is 12.47 N.

Given data:

The mass of block is m=3.37548 kg.

The accleration of block is a=3.2 m/s².

The magnitude of applied force F can be calculated by applying the Newton's second law,

`\begin{gathered} \sum ^{}_{}F_(net)=ma \\ F\cos \theta=ma \\ F\cos 30\degree=(3.37548kg)((3.2m)/(s^2))*\frac{1\text{ N}}{(1kgm)/(s^2)} \\ F=\frac{10.801536\text{ N}}{\cos 30\degree} \\ F=12.47\text{ N} \end{gathered}`

Thus, the magnitude of F is 12.47 N.