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3 years ago

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The acceleration of a particle is given by a = 6t - 28, where a is in meters per second squared and t is in seconds. Determine t

he velocity and displacement as functions of time. The initial displacement at t = 0 is s0= -8 m, and the initial velocity is v0 = 6 m/s. After you have the general expressions, evaluate these expressions at the indicated times.
Answers:

At t = 5.8 s, s = m, v = m/s
At t = 12.7 s, s = m, v = m/s

1 answer:

baherus [9]3 years ago

4 0

Answer:

Velocity, V = 3t²- 28t+6

Displacement, s = t³ - 14t²+6t -8

At t = 5.8 s

s = -249.05 m

v = -55.48 m/s

At t = 12.7 s

s = -141.48 m

v = 134.27 m/s

Explanation:

We have acceleration of a particle is given by a = 6t - 28

Velocity

$v=\int adt=\int (6t - 28)dt=3t^2-28t+C$

At t = 0 we have v₀ = 6 m/s

v₀ = 6 = 3 x 0 ²-28 x 0+C

C = 6

So velocity, V = 3t² - 28t+6

Displacement

$s=\int vdt=\int (3t^2-28t+6)dt=t^3-14t^2+6t+C$

At t = 0 we have s₀ = -8 m

s₀ = -8 = 0³ + 14 x 0² + 6 x 0 + C

C = -8

So displacement, s = t³ - 14t²+6t -8

At t = 5.8 s

s = 5.8³ - 14 x 5.8²+6 x 5.8 - 8 = -249.05 m

v = 3 x 5.8² - 28 x 5.8 + 6 = -55.48 m/s

At t = 12.7 s

s = 12.7³ - 14 x 12.7²+6 x 12.7 - 8 = -141.48 m

v = 3 x 12.7² - 28 x 12.7 + 6 = 134.27 m/s

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a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

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We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

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b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

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k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

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c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

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