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 ?

1) 2.6 N

2)

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, , downward

So the equation of the forces on block 2 is

where

is the mass

is the acceleration due to gravity

is the downward acceleration

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

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),

So the equation of the forces for block 1 is

where:

is the tension in the cord

F = 6.0 N

is the mass of the block

is the angle of the incline

is the acceleration of the block (up along the plane, so positive)

Therefore, the angle is:

gas and liquid

its a 5.5 n

glad that