Question

A child drags a 0.398 kg toy dog across flatground at constant speed, with a 4.63 N forceat a 63.0° angle. What is the component ofthe force of friction along the direction ofmotion on the toy?

Answer 1

The force acting on the toy is,

`F=4.63\text{ N}`

As the toy is moving with the constant speed,

So acceleration of the toy is zero.

According to the Newton's Law,

`\begin{gathered} F_(net)=ma \\ F_(net)=0 \end{gathered}`

Thus, the force acting in the horizontal direction is equal to the friction force acting on the toy.

`\begin{gathered} F_(friction)=-F\cos (63^(\circ)) \\ F_(friction)=-4.63*\cos (63^(\circ)) \\ F_(friction)=-2.10\text{ N} \end{gathered}`

Thus, the force of friction acting on the toy is 2.10 N in the opposite direction of the motion of the toy.