You are using a rope to lift a 14.5 kgkg crate of fruit. Initially you are lifting the crate at 0.500 m/sm/s. You then increase the tension in the rope to 180 NN and lift the crate an additional 1.15 mm. During this ddd motion, how much work is done on the crate by the tension force?

Explanation:

Initially , crate is going upwards with uniform velocity so net force on crate is zero.

Tension T = mg ( its weight )

= 14.5 x 9.8

= 142.1 N

Now tension is increased to 180 N , so net force on it

= 180 - 142.1 = 37.9 N

Work done by tension that is 180 N

= Tension x displacement

= 180 x 1.15 x 10⁻³

= .207 J

root word, prefixes, and suffixes

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.