Question

A 10.0 kg box is pulled across a horizontal surface by a 40.0 N force. The force is applied at an angle of 30.0 degrees above the horizontal. The coefficientof kinetic friction is 0.30. What is the acceleration of the box Assume the box moves in the positive direction. Answer with no units to the correctnumber of sig.fi

Answer 1

The mass of the box is given as,

`m=10\text{ kg}`

The force acting at the 30 degree is given as,

`F=40\text{ N}`

The frictional force acting on the box is,

`\begin{gathered} F^(\prime)=\mu mg \\ F^(\prime)=0.30*10*9.8 \\ F^(\prime)=29.4\text{ N} \end{gathered}`

The applied force in the horizontal direction is,

`\begin{gathered} F\cos (30^(\circ))=40*\cos (30^(\circ)) \\ =34.64\text{ N} \end{gathered}`

The net force acting on the box is,

`\begin{gathered} F_(net)=F\cos (30^(\circ))-F^(\prime) \\ F_(net)=34.64-29.4 \\ F_(net)=5.24\text{ N} \end{gathered}`

According to the newton second law,

`\begin{gathered} F_(net)=ma \\ 5.24=10* a \\ a=(5.24)/(10) \\ a=0.524ms^(-2) \end{gathered}`

Thus, the acceleration of the box is 0.524.