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 76.0 cm to the right and the 12.0-N block moves 76.0 cm downward. Required:
a. During this process, how much work is done on the 12.0-N block by gravity?
b. During this process, how much work is done on the 12.0-N block by the tension in the string?
c. During this process, how much work is done on the 20.0-N block by gravity?
d. During this process, how much work is done on the 20.0-N block by the tension in the string?
e. During this process, how much work is done on the 20.0-N block by friction?
f. During this process, how much work is done on the 20.0-N block by the normal force?

a) 9.12 Joules

b) - 9.12 Joules

c) 0

d) 9.12 Joules

e) -9.12 Joules

f) 0

Explanation:

a. During this process, how much work is done on the 12.0-N block by gravity?

Work done = Force(N) × Distance(d)

Distance = 76 cm

100 cm = 1m

76cm =

Cross Multiply

= 76cm × 1 m/ 100cm

= 0.76m

Work done = 0.76m × 12N

= 9.12 Joules

b. During this process, how much work is done on the 12.0-N block by the tension in the string?

Tension: This always occurs in an equal and opposite direction,hence:

Force = 12N, Tension = -12N

Work done = Tension(N) × Distance(d)

Distance = 76 cm

100 cm = 1m

76cm =

Cross Multiply

= 76cm × 1 m/ 100cm

= 0.76m

Work done = 0.76m × -12N

= - 9.12 Joules

c. During this process, how much work is done on the 20.0-N block by gravity?

= This 20.0N block is not moving up or down , therefore, the work done on the block by gravity is Zero = 0

d. During this process, how much work is done on the 20.0-N block by the tension in the string?

Frictional force = -12N because the Box moves in the same direction, hence:

= Tension = 12N

Work done = Tension (N) × Distance(d)

Distance = 76 cm

100 cm = 1m

76cm =

Cross Multiply

= 76cm × 1 m/ 100cm

= 0.76m

Work done = 0.76m × 12N

= +9.12 Joules

e. During this process, how much work is done on the 20.0-N block by friction?

This time around Friction is on the opposite side of Tension.

Hence:

Tension = 12N, Frictional force = -12N

Work done = Frictional force (N) × Distance(d)

Distance = 76 cm

100 cm = 1m

76cm =

Cross Multiply

= 76cm × 1 m/ 100cm

= 0.76m

Work done = 0.76m × -12N

= - 9.12 Joules

= -9.12 Joules

f. During this process, how much work is done on the 20.0-N block by the normal force?

The 20.0N block is neither moving up or down hence, the work done on the 20.0N block by the normal force is zero (0)

wheres the chart? so i can answer.

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