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1 year ago

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A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then r

eleased and allowed to oscillate in simple harmonic motion. What is the maximum acceleration of the block

1 answer:

vagabundo [1.1K]1 year ago

4 0

Answer:

x = A sin ω t describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t

a (max) = -A ω^2 is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2 for SHM

F = - K x^2 restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2

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Difference between velocity ratio and efficiency

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Explanation:

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(b) A ball is thrown from a point 1.50 m above the ground. The initial velocity is 19.5 m/s at

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The answers are:

(i) 6.35m

(ii) 20.2 m/s

It seems like you already have the answer, but let me show you how to get it:

You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)

When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*<em>Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components. </em>

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = <em>9.75 m/s</em>

Vix = CosΘVi = (Cos30°)(19.5m/s) = <em>16.89 m/s</em>

Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.

(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

Where:

Vf = final velocity

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a = 9.8m/s²

d = displacement

We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

$dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

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Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$

$Vfy^{2} = 0m/s^{2}+2(9.8m/s^{2}(6.35m)$

$\sqrt{Vfy^{2}} = \sqrt{124.46m^{2}/s^{2}}$

$Vfy= 11.16m/s$

OKAY! We are at our last step. Now to get the resultant velocity, we apply the Pythagorean theorem.

$Vf^{2} = Vfx^{2} + Vfy^{2}$

$\sqrt{Vf^{2}} = \sqrt{(16.89m/s)^{2}+(11.16m/s)^{2}}$

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Answer:

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