Question

Can someone help with this

Answer 1

Hi there!

A.

We can use the work-energy theorem to solve.

Initially, both blocks have gravitational potential energy, which is then converted to kinetic energy.

Recall:
$U = mgh$

U = Potential Energy (J)
m = mass (kg)
g = acceleration due to gravity (m/s²)

h = height (m)

And at the end, block 2 has both kinetic and potential energy. Kinetic is defined as:
$KE = \frac{1}{2}mv^2$

v = velocity (m/s)

We can do a summation of initial and final forces.

Initial:
$E_i = m_2g\frac{h}{2} + m_1gh$

Final:
$E_f = \frac{1}{2}m_2 v^2 + m_2g\frac{3h}{2} + \frac{1}{2}m_1 v^2$

Set the two equal (no energy loss in this situation) and solve for velocity.

$m_2g\frac{h}{2} + m_1gh = \frac{1}{2}m_2 v^2 + m_2g\frac{3h}{2} + \frac{1}{2}m_1 v^2\\\\m_2g\frac{h}{2} + m_1gh - m_2g\frac{3h}{2} = \frac{1}{2}m_2 v^2 + \frac{1}{2}m_1 v^2\\\\v^2 = \frac{gh(\frac{m_2}{2} + m_1 - \frac{3m_2}{2})}{\frac{1}{2}m_2 + \frac{1}{2}m_1}\\\\v = \sqrt{\frac{gh(\frac{m_2}{2} + m_1 - \frac{3m_2}{2})}{\frac{1}{2}m_2 + \frac{1}{2}m_1}}\\\\v = \boxed{3.704 \frac{m}{s}}$

**This can be solved in a simpler way using a summation of forces.

B.

Now, we can do a summation of forces to solve. First, we must solve for the acceleration of the system now that we have found the velocity. Use the kinematic equation:
$v_f^2 = v_i^2 + 2ad$

Initial velocity is 0 m/s, so:
$v_f^2 = 2ad\\\\a = \frac{v_f^2}{2d} = \frac{(3.704^2)}{2(2.10)} = 3.267 \frac{m}{s^2}$

Now, we can use a summation of forces for any block. We can do block 1:
$\Sigma F = m_1g - T$

Using Newton's Second Law:
$m_1a = m_1g - T\\\\T = m_1g - m_1a \\T = m_1 (g - a) = 4.10(9.8 - 3.267) = \boxed{26.828 N}$