Question

A 15.0 kg load of bricks hangs from one end of a rope that passes over a small, frictionles pulley. A 28.0 kg counterweight is suspended from the other end of the rope . the system is released from rest. A0 draw two free-body diagrams, one for the load od bricks and one for the counterweight. B) What is the magnitude of the upward acceleration of the load of bricks

Answer 1

Answer:

A) The free body diagrams for both the load of bricks and the counterweight are attached.

B) a = 2.96 m/s²

Explanation:

A)

The free body diagrams for both the load of bricks and the counterweight are attached.

B)

The acceleration of upward acceleration of the load of bricks is given by the following formula:

a = g(m₁ - m₂)/(m₁ + m₂)

where,

a = upward acceleration of load of bricks = ?

g = 9.8 m/s²

m₁ = heavier mass = mass of counterweight = 28 kg

m₂ = lighter mass = mass of load of bricks = 15 kg

Therefore, using these values in equation, we get:

a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)

a = 2.96 m/s²

Answer 2

Answer:

2.96 m/s2

Explanation:

Set Up: Apply ∑→ Fy=m → ay to the load of bricks & to the counterweight.  

The tension is the same at each end of the rope.  

The rope pulls up with the same force T on the bricks & on the counterweight.    

The counterweight accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude.  

Solve: Apply ∑→ Fy=m → ay to each object. The acceleration magnitude is the same for the two objects.  

For the bricks take +y to be upward since vector a (→ a) for the bricks is upward: ∑→ Fy=m → ay  

T – m1g = m1a  

For the counterweight take +y to be downward since → a is downward: ∑→ Fy=m → ay ,  m2g – T = m2a  

Add the two equations to eliminate T and then solve for a.  

m2g ⎯ m1g = m1a + m2a  

(m2⎯ m1)g = (m1 + m2)a  

a = (m2− m1/ m1 + m2)g    

  = (28 kg−15 kg/15 kg+28 kg) (9.8 m/s2) = 2.96 m/s2