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devlian [24]
3 years ago
12

A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the e

quilibrium point (y= 0) with a negative velocity at t= 0.Assume that downward is the positive direction of motion. What equation describes this motion as a function of time?
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

 y = -19.2 sin (23.15t) cm

Explanation:

The spring mass system is an oscillatory movement that is described by the equation

      y = yo cos (wt + φ)

Let's look for the terms of this equation the amplitude I

     y₀ = 19.2 cm

Angular velocity is

     w = √ (k / m)

     w = √ (245 / 0.457

     w = 23.15 rad / s

The φ phase is determined for the initial condition   t = 0 s ,  the velocity is negative v (0) = -vo

The speed of the equation is obtained by the derivative with respect to time

     v = dy / dt

     v = - y₀ w sin (wt + φ)

For t = 0

     -vo = -yo w sin φ

The angular and linear velocity are related v = w r

      v₀ = w r₀

      v₀ = v₀ sinφ

      sinφ = 1

      φ = sin⁻¹ 1

      φ = π / 4    rad

Let's build the equation

      y = 19.2 cos (23.15 t + π/ 4)

Let's use the trigonometric ratio π/ 4 = 90º

      Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a

       y = -19.2 sin (23.15t) cm

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Explanation:

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A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.
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Refer to the figure shown below.

g =  9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
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For static equilibrium,
μN = F
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μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
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0.9563μW > 0.2924W
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Answer:
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The minimum coefficient of static friction is 0.306

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pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

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60 J = <em>m g h</em>

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