Respuesta :
Answer:
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)
