91Ó°ÊÓ

• Mass of the car = 1100 kg.
• Force exerted by the car = 1900 N.
• Mass of the boat and trailer = 700 kg.
• Acceleration = 0.550 m/s².

Apply Newton’s second law of motion as:

$\begin{array}{c}{\mathit{F}}_{\mathit{n}\mathit{e}\mathit{t}}\mathbf{=}\mathit{M}\mathit{a}\\ \mathit{F}\mathbf{−}\mathit{f}\mathbf{=}\mathbf{(}{\mathit{m}}_{\mathit{b}}\mathbf{+}{\mathit{m}}_{\mathit{t}}\mathbf{+}{\mathit{m}}_{\mathit{c}}\mathbf{)}\mathit{a}\end{array}$ …â¶Ä¦â¶Ä¦â¶Ä¦.…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(¾±)

Here, F_net is the net force acting on the system, m_b is the mass of the boat, m_tis the mass of the trailer, m_c is the mass of the car, a is the acceleration, and F is the force exerted by the car, and f is the resistance force.

Substitute 700 kg for (m_b+m_t), 1100 kg form_c, 1900 N forF, and 0.550 m/s² for a in the above expression, and we get,

$\begin{array}{c}1900\text{ N}−f=\left(700+1100\right)\text{ k²µ}×{\text{0.55 m/²õ}}^2\\ 1900\text{N}−f=1800\text{ k²µ}×{\text{0.55 m/²õ}}^2\\ f=\left(1900−990\right)\text{ N}\\ f=910\text{ N}\end{array}$

Hence, the total resistance force is 910 N.

Apply Newton’s second law of motion as:

$\mathit{F}\mathbf{\text{'}}\mathbf{−}\mathbf{0}\mathbf{.8}\mathit{f}\mathbf{=}\left(m_b+m_t\right)\mathit{a}$

Here, F’ is the force in the hitch between the car and the trailer.

Substitute 900 N forf, 700 kg for (m_b+m_t), and 0.550 m/s² for a in the above expression, and we get,

localid="1655713865611" $\begin{array}{c}F\text{'}−0.8×910\text{ N}=700\text{ k²µ}×{\text{0.55 m/²õ}}^2\\ F\text{'}−728\text{ N}=\text{385 N}\\ \mathit{\text{F'}}=\left(385+728\right)\text{ N}\\ \mathit{\text{F'}}=1113\text{ N}\end{array}$

Hence, the force in the hitch between the car and the trailer is 1113 N.