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frozen [14]
3 years ago
13

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 the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.
Physics
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

a. t=0.5s

b. f=2Hz

c. w =4π

d. k =  78.95 N/m

e. v = 4.39 m/s

f. Fs=4.835N

Explanation:

a. The period of the motion is

t= 0.5 s

b. Frequency

f =1 / t

f = 1 / 0.5 = 2 Hz

c. Angular frequency

w = 2π*f

w = 2π*2Hz= 4π

d. The spring constant is

f = 1/2π * √k/m

k = (2π * f)^2*m =(2π*2Hz)^2 * 0.5 kg

k =  78.95 N/m

e. Maximum speed

Kinetic energy and spring energy to find the maximum speed

1/2 *K*x^2=1/2*m*v^2

1/2 * 78.95N/m*(0.35m)^2=1/2*0.5kg*v^2

Solve to v'

v = 4.39 m/s

f. The maximum force on the block from the spring

Fs= 1/2*K*x^2

Fs=1/2*78.95N/m*(0.35m)^2

Fs=4.835N

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Answer:

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On rearranging the above equation, to find the angle as :

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A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at
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Answer:

 μ = 0.18

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Answer:

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Explanation:

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the equation given in the problem is

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          λ = 2π / 1.65

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Answer:

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Explanation:

a)

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       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
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  • Replacing (2) and (3) in (1), we can solve for α, as follows:

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  • Since the angular acceleration is constant, we can use the following kinematic equation:

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  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
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b)    

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       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

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