Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of the block as it passes through the equilibrium position is 62 cm/s. What is the angular frequency of the block's motion
1 answer:
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
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Answer:
Velocity = 4.33[m/s]
Explanation:
The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.
All this energy will become kinetic energy and we can find the velocity.
To find speed you have to divide distance by time. In this case:
Explanation:
a) How much work is done by gravity?
w = f x d w = 950 x 10 x 5.5 = 52250j b) How much work is done by tension?
v²=u²+2as 0.75²=0.25²+2a x5.5 0.56=0.06+2a x5.5 2a x5.5 = 0.56 - 0.06 2a x 5.5 =0.5 11a=0.5 a = 0.5/11 = 0.05m/s² w = f x d
w = 950 x 0.05 x 5.5 = 261.25j
