Question

A barbell of mass 5 kg is raised to a height of 7m above the ground.Calculate the work done by the boy.Take acceleration due to gravity g = 10 ms? .​

Answer 1

Answer:

Explanation:

Work against the gravity will be PE=mgh=5×10×10=500J

Total work done by the external force is W=Fd=170×10=1700J  

so the kinetic energy gain will be KE=W−PE=1700J−500J=1200J

Answer 2

Answer:

#rajat20112

Explanation:

Work against the gravity will be PE=mgh=5×10×10=500J

Total work done by the external force is W=Fd=170×10=1700J

so the kinetic energy gain will be KE=W−PE=1700J−500J=1200J

if the velocity becomes v then

2

1

mv

2

=1200Joule

or velocity v=

5

2400

=21.9m/s

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