A block weighing 17 N oscillates at one end of a vertical spring for which k = 120 N/m; the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched 0.24 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?
No the amplitude is not 0.24 m read the question carefully, u will notice that 0.24 m is beyond its relaxed length (the length when no object is attached
so the correct amplitude will be
0.24m-(17N/120N/m) = 0.0983m
If v=0, the block is at its extreme (end) position =>
x= x(max)=0.24 m = A (amplitude)
The equation of the oscillation is
x=Asinωt
v=Aωcosωt
a= - Aω²sinωt
F= - m Aω²sinωt = - mω²x = - kx,
x=A => F= kA = 120∙0.24= 28.8 N.
ω= √(k/m)= √(kg/W)=
=√(120∙9.8/17) = 8.32 rad/s
T=2π/ ω=2π/8.32=0.76 s.
KE(max) =PE(max) =kA²/2 = 120∙0.24²/2 = 3.456 J.
To answer this question, we will use the principles of simple harmonic motion.
(a) The net force on the block can be determined by calculating the force exerted by the spring at the given displacement. According to Hooke's Law, the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The formula for Hooke's Law is: F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, k = 120 N/m and x = 0.24 m. Plugging these values into the formula, we get:
F = -(120 N/m)(0.24 m) = -28.8 N
The negative sign indicates that the force is directed in the opposite direction of the displacement, which means it is directed upward.
Therefore, the net force on the block at this instant is 28.8 N directed upward.
(b) The amplitude of the simple harmonic motion is equal to the maximum displacement from the equilibrium position. In this case, the spring is stretched 0.24 m beyond its relaxed length, so the amplitude is 0.24 m.
(c) The period of simple harmonic motion can be calculated using the formula:
T = 2π√(m/k)
where T is the period, m is the mass of the block, and k is the spring constant.
However, the mass of the block is not given in the question. To find the period, we need to determine the mass. We can do this by using the formula:
F = mg
where F is the force of gravity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the weight of the block is 17 N, we have:
17 N = mg
Rearranging this formula, we can solve for m:
m = 17 N / 9.8 m/s^2
m ≈ 1.73 kg
Now we can calculate the period using the formula:
T = 2π√(m/k)
T = 2π√(1.73 kg / 120 N/m)
T ≈ 2.51 s
Therefore, the period of the simple harmonic motion is approximately 2.51 seconds.
(d) The maximum kinetic energy of the block can be determined by using the formula:
Kmax = (1/2)mv^2
where Kmax is the maximum kinetic energy, m is the mass of the block, and v is the maximum velocity.
In simple harmonic motion, the maximum velocity and amplitude are related by the equation: vmax = ωA
where vmax is the maximum velocity, ω is the angular frequency, and A is the amplitude.
The angular frequency can be calculated using the formula: ω = √(k/m)
Substituting the given values, we have:
ω = √(120 N/m / 1.73 kg)
ω ≈ 7.45 rad/s
Now we can calculate the maximum velocity:
vmax = ωA
vmax ≈ (7.45 rad/s) * (0.24 m)
vmax ≈ 1.79 m/s
Finally, we can calculate the maximum kinetic energy:
Kmax = (1/2)mv^2
Kmax = (1/2)(1.73 kg)(1.79 m/s)^2
Kmax ≈ 2.06 J
Therefore, the maximum kinetic energy of the block as it oscillates is approximately 2.06 Joules.