Question

E. O 1.318 m/s²6. An object of mass 32 kg is being pulled on a friction lesshorizontal surface by means of two strings. One of the stringsis pulling forward, has a tension of 120 N and makesan angle of 30 degree with the horizontal-right. The other string ispulling backwards horizontally and has a tension of 10 N.Calculate the acceleration of the object. (1 point)A. 02.935 m/s²B.O2.506m/s2C. 05.172 m/sD. 03.816 m/s ²E. O4.308 m/s²

Answer 1

We first determine the total force ni he horizontal axis, that is:

`\sum F_x=120cos(30)N-10N\Rightarrow\sum F_x=(60√(3)-10)N`

Now, we determine the acceleration:

`\begin{gathered} (60√(3)-10)N=(32kg)a\Rightarrow a=((60√(3)-10)N)/(32kg) \\ \\ \Rightarrow a=(30√(3)-5)/(16)m/s^2\Rightarrow a\approx2.935m/s^2 \end{gathered}`

So, the acceleration is approximately 2.935 m/s^2.