Two crates, one with mass 4.00 kgand the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (Fig. P4.39). A woman wearing golf shoes (for traction) pulls horizontally on the 6.00-kgcrate with a force Fthat gives the crate an acceleration of $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a) What is the acceleration of the 4.00 kgcrate? (b) Draw a free-body diagram for the 4.00 kgcrate. Use that diagram and Newton’s second law to find the tensionin the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00 kgcrate. What is the direction of the net force on the 6.00-kgcrate? Which is larger in magnitude,Tor F? (d) Use part (c) and Newton’s second law to calculate the magnitude of F.

The given data is listed below as:

• The mass of the first crate is,$m_a=4.00-kg$
• The mass of the second crate is,${\mathrm{m}}_2=6.00-\mathrm{kg}$
• The acceleration of the crate is,$\mathrm{a}=2.50\mathrm{m}/{\mathrm{s}}^2$

The summation of the forces is described as the product of the forces with the help of the sequential movement of the parts. Moreover, the summation of the forces mainly produces the optimum amount.

As both the crates are being pulled together, then the acceleration of the crate will not change. Hence, the acceleration of the crate will remain same.

Thus, the acceleration of the crate is $2.50\mathrm{m}/{\mathrm{s}}^2$.

The free body diagram for the first crate has been drawn below:

In the above diagram, the tension T is acting in the left direction of the crate; the weight of the crate $w_2$is acting downwards. The normal force acting on the crate is$n_2$ and the force acting due to the tension isF that is the product of the mass of the crate$m_1$ and acceleration due to gravity a.

From the above diagram, the summation of the forces acting in thex direction is zero.

The equation of the summation of the forces acting in thex direction is expressed as:

$T=m_{1}a$

Here,T is the tension, $m_1$ is the mass of the first crate and a is the acceleration of the crate.

Substitute 4.00-kg for $m_1$and $2.50m/s^2$ for a in the above equation.

$$\begin{gathered} \mathrm{T}=\left(4.00-\mathrm{kg}\right){\left(2.50\mathrm{m}/\mathrm{s}\right)}^2 \\ =10\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =10\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =10\mathrm{N} \end{gathered}$$

Thus, the tension in the rope is 10 N.

The free body diagram for the second crate has been drawn below:

In the above diagram, the tensionT is acting in the left direction of the crate; the weight of the crate$w_1$ is acting downwards. The normal force acting on the crate is$n_1$ and the force acting due to the tension is F. Another force is acting on the right side of the crate that is the product of the mass of the second crate$m_2$ and acceleration due to gravity a.

According to the above diagram, the net force is acting at the right direction of the block that is in the positive direction of the x axis. The reason that the force is acting in that direction is that it will help to accelerate the crate.

Thus, the net direction of the force is in the positive direction of the x axis.

According to the above diagram, the force is equal to the addition of the tension and the product of the mass of the second crate and the acceleration. Hence, the force F is larger than the tension T.

Thus, the force F is larger than the tension T.

From the above diagram, the summation of the forces acting in the x direction is zero.

The equation of the summation of the forces acting in the x direction is expressed as:

$F=T+m_{2}a$

Here, T is the tension, F is the force of the crate, $m_2$ is the mass of the second crate and a is the acceleration of the crate.

Substitute 6.00-kg for $m_1$, 10 N forT and $2.50\mathrm{m}/{\mathrm{s}}^2$forain the above equation.

$$\begin{gathered} \mathrm{F}=\left(10\mathrm{N}\right)+\left(6.00\mathrm{kg}\right)\left(2.50\mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(10\mathrm{N}\right)+15\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =\left(10\mathrm{N}\right)+15\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =25\mathrm{N} \end{gathered}$$

Thus, the magnitude of the force is 25 N.