Question

A cord is used to lower vertically a block of mass M, through
a distance d at a constant downward acceleration of g/8.
Then the work done by the cord on the block is
(a) Mg d/8 (b) 3 Mg d/8
(c) Mg d (d) – 7 mg d/8

Answer 1

Answer:

b..

Explanation:

Hope this helps you..

Plz mark it as brainliest ❣️.

Answer 2

Answer:

• The work done by the chord on the block is - 7 Mg/8

Explanation:

Given that:-

• The mass of the block is denoted by 'M'
• It moves a distance 'd' in downward direction
• it moves downward with a constant acceleration of g/8

To Find:-

• The work done by the chord on the block

Formula used:-

• Net force = ma
• work done = FS Cos∅

Terms to remember:-

• Tension is always a pulling force
• The weight ( Mg ) of the block acts down ward
• The angle b/w the force ang displacement is 180 degrees since they are opposite in direction

Required Solution:-

• Since the block is lowered downward there would be tension developed in the chord acting upward

List of force acting up on the block  :

→ The tension developed in the chord upward ( T )

→ The weight of the block downward ( Mg )

• We know that block moves downward with a acceleration g/8

Thereby:-

• The downward force is greater than that of the upward force

Framing an equation :-

→ F ( Net ) = Ma

→ Mg - T = Mg/8

→ Mg - T = Mg - 7 M/g

Comparing both the sides :-

• Tensional force = 7 Mg/8

Work done by the chord:-

→ work done by the chord = Fscos∅

→ Work done by the chord = 7 Mg/8 × d × cos ( 180 )

→ Work done by the chord = 7 Mg/8 × d × - 1

→ Work done by the chord = - 7 Mg/8 d

Therefore:-

• The work done by the chord is - 7 Mgd/8  

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬