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KonstantinChe [14]
3 years ago
9

At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the

spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. At what time will the block next pass x = 0, the place where the spring is unstretched? t1 =_______.
Physics
1 answer:
Ber [7]3 years ago
7 0

Answer:

0·149 s

Explanation:

Initially at t = 0 s, velocity of the block = v = 2 m/s and x_{0} = - 0·33 m from the equilibrium position of the spring

Spring constant k = 61·2 N/m

Acceleration of the block at any instant of time = - (k × x) ÷ m

<h3>As acceleration of the block depends on x and acts in opposite direction to the motion of the block</h3><h3>The motion of the block will be simple harmonic motion</h3><h3>Then the equation of motion = x = A × sin(w × t + c)</h3>

Where x is the distance of the block from equilibrium position

A is the amplitude( that is maximum distance from the equilibrium position)

w is the angular frequency

t is the time taken

c is any constant

<h3>For simple harmonic motion a = - w² × x</h3>

- w² × x =  - (k × x) ÷ m

From the above equation w = √(k ÷ m) = √(61·2 ÷ 5) = 3·499 rad/s

By substituting the values in the given equation

- 0·33 = A × sin(c) → equation 1

By differentiating the equation x = A × sin(w × t + c) with respect to t on both sides

v = A × w × cos(w × t + c)

By substituting the values

2 = A × w × cos(c) → equation 2

By dividing the equation 1 and equation 2

- (0·33 ÷ 2) × w = tan(c)

tan(c) = - 0·577

⇒ c = π - inverse of tan(0·577)

∴ c = π - 0·523 rad

Substituting the value of c in equation 1

- 0·33 = A × sin(π - 0·523)

∴ A = - 36·67 m

∴ x = - 36·67 × sin(3·499 × t + π - 0·523)

At x = 0

sin(3·499 × t + π - 0·523) = 0

∴ 3·499 × t + π - 0·523 = 0 or π

It can't be 0 because if it is 0, then t is negative

∴ 3·499 × t + π - 0·523 = π

3·499 × t = 0·523

∴ t = 0·523 ÷ 3·499 = 0·149 s

∴ Here t_{1} = 0·149 s

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<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

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<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

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v is the final speed

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a = g = 9.8 m/s^2 is the acceleration of gravity

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Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

d = 1.8 mm = 0.0018 m

So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

Solving for a,

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

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a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

a=\frac{v'-v}{t}

where

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For the hardwood floor,

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