A block of mass 3.31 kg is hanging from a rope. The tension in the rope is 18.4 N, pulling upward on the block. What is the magnitude and direction of the acceleration of the block?

Question

A block of mass 3.31 kg is hanging from a rope. The tension in the rope is 18.4 N, pulling upward on the block. What is the magnitude and direction of the acceleration of the block?

Answer 1

Solution:

The weight of the block pulling down is W = mg , W = 3.31 kg ( 9.81 m/s²) = 32.5 N ↓

The force in the rope is the tension up = F = 18.4 N ↑

If you cut the rope and have a free body diagram, the net force is the sum of the W and F

The sum of forces ΣF = ma , and ΣF = 32.5 N ↓ + 18.4 N ↑ considering the downward position as negative direction ΣF = -32.4 N + 18.4 N = -14.1 N ↓ and negative

The acceleration a = ΣF / m = -14.1 N ↓ / 3.31 kg = (-14.1 kg.m/ s²) / 3.31 kg = -4.26 m/s² down ↓ and negative

Answer 2

Use the equations

∑F_x = ma_x

∑F_y = ma_y

There are no forces on the block in the x-direction (horizontally) so we don't need to use the first equation.

From the second equation, we have two forces on the block in the y-direction.

1) The gravitational force (the weight of the block) going down

2) The tension going up

Sum up these forces:

F_weight + F_Tension = ma_y

3.31kg(-9.81m/s²) + 18.4N = 3.31kg(a_y)

Solve for the acceleration in the vertical direction