A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block
1 answer:
Answer:
+ 24 N
Explanation:
the computation is shown below:
Given that
Mass of the block = m = 0.7 kg
Sprint constant = k = 160 N / m
x = 0.15m
Now the force on the block is
F = kx
= (160) (0.15)
= 24 N
As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction
Therefore the third option is correct
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Answer:
Explanation:
f =
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = )
f = /6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N