1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
Answer:

Explanation:
Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

Due to the definition of cross product, the magnitude of the torque is given by:

Where
is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when
is equal to one, solving for r:

Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J