Two blocks are connected by a very light string passing over a massless and frictionless pulley. Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string

Explanation:

Please see attached the diagram and also the FBD for your reference.

Step one:

given data

Weight w1=20N

Distance s1=75cm= 0.75m

Weight w2=12N

Distance s2=75cm =0.75m

Clearly, we are asked to find the work done by gravity and the tension in the string.

we know that work is defined as force times the distance traveled

W=F*D

and to know the force we need the acceleration, seeing that the system is stationary the acceleration due to gravity 9.81m/s^2 is not needed here

i . Work done by gravity = w2*s2=12*0.75=9J

ii. The tension is equal but opposite = -9J

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