Question

A block of mass 6 kg is lowered with the help of a rope of negligible mass through a distance of 20 m with an acceleration of 1.96 meters second square work done by the rope on the block is

Answer 1

Answer:235.2J

Explanation:

Work done= F×s

F=ma

=6×1.96

=11.76

W=F×S

=11.76×20

=235.2J

Answer 2

Given:

Mass of the block = 6kg

Mass of the rope = 20m

Acceleration = 1.96m

To Find:

Square work done on the rope.

Solution:

Work done = F = ma

= 6 × 1.96

= 11.76

To calculate square work done -  

W = f × s

= 11.76 × 20

= 235.2

Answer: The work done is 235.2J