How Can Physics Solve These Complex Problems?

[real2]

1. An iron anchor of density 7870 kg/m3 appears 200 N lighter in water than in air. (a) What is the volume of the anchor? (b) How much does it weigh in air?


2. An oscillator consists of a block of mass 0.500 kg connected to a spring. When set into oscillation with amplitude 35.0 cm, the oscillator repeats its motion every 0.500 s. Find (a) the period, (b) the frequency, (c) the angular frequency, (d) the spring constant, (e) the maximum speed, and (f) the magnitude of the maximum force on the block from the spring.


3. A 10 g particle is undergoing simple harmonic motion with amplitude 2.0×10–3m and a maximum acceleration of magnitude 8.0×10–3 m/s2. The phase constant is –π/3rad. (a) Write an equation for the force on the particle as a function of time. (b) What is the period of the motion? (c) What is the maximum speed of the particle? (d) What is the total mechanical energy of this simple harmonic oscillator?

 

[KoGs]

Show your attempts at solving it or no one can help you.

 

[kyle8921]

1. (a) To find the volume of the anchor, we can use the equation for density: density = mass/volume. Rearranging the equation, we get volume = mass/density. Therefore, the volume of the anchor is 200 N/7870 kg/m3 = 0.0254 m3.

(b) To find the weight of the anchor in air, we can use the equation for weight: weight = mass*gravity. Rearranging the equation, we get mass = weight/gravity. Therefore, the weight of the anchor in air is 200 N/9.8 m/s2 = 20.4 kg.

2. (a) The period of the oscillator can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Plugging in the given values, we get T = 2π√(0.500 kg/k) = 2.52 s.

(b) The frequency can be found using the formula f = 1/T, where f is the frequency and T is the period. Substituting the value of T from part (a), we get f = 1/2.52 s = 0.397 Hz.

(c) The angular frequency can be calculated using the formula ω = 2πf, where ω is the angular frequency and f is the frequency. Substituting the value of f from part (b), we get ω = 2π*0.397 Hz = 2.49 rad/s.

(d) The spring constant can be found using the formula k = mω2, where k is the spring constant, m is the mass, and ω is the angular frequency. Substituting the given values, we get k = (0.500 kg)*(2.49 rad/s)2 = 3.10 N/m.

(e) The maximum speed of the block can be calculated using the formula vmax = Aω, where vmax is the maximum speed, A is the amplitude, and ω is the angular frequency. Substituting the given values, we get vmax = (0.350 m)*(2.49 rad/s) = 0.873 m/s.

(f) The magnitude of the maximum force on the block from the spring can be found using the formula Fmax = kA, where Fmax