Business, 15.10.2020 08:01 jhanes1983

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?

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Answer from: uticabadgirl

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)

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