Question

1. Ignore friction and determine an expression for the distance d the boxes travel before coming to rest. Choose the floor as the position for zero gravitational potential energy. (Use the following as necessary: m1, m2, v, and g for the acceleration due to gravity.)
2. Determine an expression for the work done on box 1 by the rope. (Use the following as necessary: m1, m2, v, and g for the acceleration due to gravity.)

Answer 1

Newton's second law allows us to find the results for the displacement and work on box 1 are:

       1) The displacement is   $x = \frac{v_o^2 m_2}{2(m_1+m_2)} g$

      2) The work is  $W= \frac{1}{2} m_1v_o^2$

1) Newton's second law says that force is directly proportional to the mass and acceleration of bodies.

         F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the bodies

The reference system is a coordinate system with respect to which the decompositions of the forces are carried out and all the measurements are made, in this case we take a system where the x axis is horizontal, the y axis is vertical and the zero of the system is located at soil

In the attached we can see a free body diagram of the system where the external forces are indicated, let's apply Newton's second law to body 1

body 1

x-axis

          -T = m₁ a

body 2

 y-axis

           T - W₂ = m₂ a

           W₂ = m₂ g

let's solve the system

        - m₂ g = (m₁ + m₂) a

           a = $- \frac{m_2}{m_1+m_2} \ g$- m2 / m1 + m2 g

Kinematics studies the movement of bodies, let's use the expression

            v² = $v_o^2$ - 2 a x

When the body stops the velocity is zero

            x = $\frac{v_o^2}{2a}$

We substitute

            $x = \frac{v_o^2}{2} \frac{m_2}{m_1+m_2} \ g$

Since the two boxes are connected by a rope, they both travel the same distance

           

2) They ask for Tension work on box 1

Work is defined by the scalar product of force and displacement

        W = F. d

Where the bold letters indicate vectors, W is the work that is a scalar, F the force and d the displacement

We can write this expression by developing the dot product

       W  = F d cos θ

Where θ is the angle between force and displacement.

In this case the force is the tension of the rope, from the attached graph we see that the force is directed to the right and the displacement is to the left, therefore the angle is 180º and the cos  180 is equal to -1

We look for the tension from Newton's second law for box 1

          T = - m₁ a

Let's substitute

          T =$- m_1 \ ( - \frac{m_2}{m_1+m_2} ) \ g$  

          T = $\frac{m_1m_2}{m_1+m_2} \ g$

         

We calculate the work

         $W = - \frac{m_1m_2}{m_1+m_2 } \ g ( \frac{v_o^2 (m1+m_2)}{2 m_2 g} )$

         W = - ½ m₁ v₀²

In conclusion using Newton's second law we can find the results for the displacement and work on box 1 are:

1)  The displacement is  x= $\frac{v_o^2 m_2}{2(m_1+m_2)} g$

2) The work is  W = - ½ m₁ v₀²

Learn more here: brainly.com/question/17290735