Question

Block 1 (m1 = 1.0 kg) sits on a frictionless inclined surface and is connected to block 2 (m2 = 2.0 kg). The pulley is massless and frictionless. An upward force of magnitude F = 6.0 N acts on block 2, which has a downward acceleration of 5.5 m/s2 . What is (a) the tension in the connecting cord and (b) the angle ?

Answer 1

1) 2.6 N

2) $18.4^{\circ}$

Explanation:

1)

The force acting on block 2, which is the one hanging vertically, are:

- The upward force of magnitude F = 6.0 N

- The tension in the rope, T, upward

- The force of gravity, $m_2 g$, downward

So the equation of the forces on block 2 is

$F+T-m_2 g = m_2 a$

where

$m_2 = 2.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity

$a=-5.5 m/s^2$ is the downward acceleration

Solving the equation for T, we find the tension in the cord:

$T=m_2 a +m_2 g-F=(2.0)(-5.5)+(2.0)(9.8)-6.0=2.6 N$

2)

For this part instead, we analyze the situation for block 1.

Block 1 sits on the frictionless inclined surface, connected to block 2 via the rope.

Therefore, the forces acting on block 1 along the inclined plane are:

- The tension in the cord, T, forward (up along the plane)

- The force F, acting in the same direction of T

- The component of the weight parallel to the plane, backward (down along the plane), $m_1g sin \theta$

So the equation of the forces for block 1 is

$T+F-m_1 g sin \theta = m_1 a$

where:

$T=2.6 N$ is the tension in the cord

F = 6.0 N

$m_1=1.0 kg$ is the mass of the block

$\theta$ is the angle of the incline

$a=5.5 m/s^2$ is the acceleration of the block (up along the plane, so positive)

Therefore, the angle is:

$sin \theta = \frac{T+F-m_1 a}{m_1 g}=\frac{2.6+6.0-(1.0)(5.5)}{(1.0)(9.8)}=0.316$

$\theta=sin^{-1}(0.316)=18.4^{\circ}$