91Ó°ÊÓ

• The mass of the girl is,$m=40kg.$
• The mass of the sled is,$m=8.4kg.$
• The horizontal distance between the girl and sled is,$x=15m.$
• The horizontal force exerted on the rope by girl is,$F=5.2N.$

Newton’s second law states that force on the body is equal to the product of mass and acceleration. As per Newton’s third law, every action has an equal and opposite reaction.

To solve this problem, we need to draw a free-body diagram. By using Newton’s second law and third law we can find the acceleration of sled and girl. By using the second kinematical equation we can find the time at which the girl and sled to meet each other.

Formulae:

According to the Newton’s second law, the force acting on the sled is,

$F_s=m_s{a}_s$

Then acceleration of the sled is,

role="math" localid="1657166951978" $$\begin{gathered} a_s=\frac{F_s}{m_s} \\ =\frac{5.2\mathrm{N}}{8.4\mathrm{kg}} \\ =0.62\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the magnitude of acceleration of the sled is$0.62\mathrm{m}/{\mathrm{s}}^2$

According to Newton’s second law, the force acting on the girl is,

$F_g=m_g{a}_g$

According to the third law, the force acting on the girl is the same as the force on the sled but in opposite direction.

Then acceleration of the girl is,

$$\begin{gathered} a_g=\frac{F_g}{m_g} \\ =\frac{5.2N}{40kg} \\ =0.13\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the magnitude of acceleration of the girl is$0.13\mathrm{m}/{\mathrm{s}}^2$.

Consider the motion of the girl along positive x axis. She starts from the origin hence$x_0=0m$. The initial speed of her is $v_0=0m/s$Hence, we can use second kinematical equation for girl is,

$$\begin{gathered} x_g=x_0+v_{0}t+\frac{1}{2}{a}_g{t}^2 \\ {x}_g=\frac{1}{2}{a}_g{t}^2-\left(1\right) \end{gathered}$$

Now consider the sled is moving along negative x axis with initial position as $x_0=15m$.

The initial speed of it is$v_0=0m/s$. Hence, we can use second kinematical equation for sled is,

role="math" localid="1657167973231" $$\begin{gathered} x_s=x_0+v_{0}t+\frac{1}{2}{a}_s{t}^2 \\ {x}_s=x_0-\frac{1}{2}{a}_s{t}^2 \end{gathered}$$

When they meet, then $x_g=x_s$

$$\begin{gathered} \frac{1}{2}{a}_g{t}^2=x_0-\frac{1}{2}{a}_s{t}^2 \\ {t}^2=\frac{2x_0}{a_g+a_s}-\left(2\right) \end{gathered}$$

We have to find the position at which they meet, hence from origin it will be$x_g$

We use equation (2) in equation (1) as

$$\begin{gathered} x_g=\frac{1}{2}{a}_g\left(\frac{2x_0}{a_g+a_s}\right) \\ =\frac{2_g{x}_0}{a_g+a_s} \\ =\frac{0.13m/s^2×15m}{0.13m/s^2+0.62m/s^2} \\ =2.6m \end{gathered}$$

Hence, the position of the girl where they meet to each other is$x_g=2.6m$.