The work done by force on a spring hung from the ceiling will be 1.67 J
Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.
Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.
We need to find the work done by the force
Given mass is of 2 kg
So let,
F = 2 kg
x = 0.1 m
Stiffness of spring = k = F/x
k = 20/0.006 = 333 n/m
Now the formula to find the work done by force will be as follow:
Workdone = W = 0.5kx²
W = 0.5 x 333 x 0.1²
W = 1.67 J
Hence the work done by force on a spring hung from the ceiling will be 1.67 J
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Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
Explanation:
130km = (0.621 * 130) mi = 80.73 mi
So if you drive at 80.73 mph you will get a ticket for speeding over the limit.
Answer:
2 m/s
Explanation:
The total time = 1 hour
The vertical displacement = 1 - 1
Vertical displacement = 0
Horizontal displacement = 4 - 2
Horizontal displacement = 2
Total displacement = sqrt (2^2 - 0^2)
Displacement - 2
Average velocity is displacement/time
= 2x1
= 2 m/s
The average velocity is 2 metres per second.
