Physics, 21.07.2019 15:30 brooklym2940

A0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at (e) the

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There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
$E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
$E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
$E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
$\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
$v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
$E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
E=U+K

we find the kinetic energy when the glider is at this position:
K=E-U=0.36 J - 0.05 J = 0.31 J

And then we can find the corresponding velocity:
$K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
$a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2

e) the total mechanical energy of the glider at any point in its motion.

we have already calculated it at point b), and it is given by
$E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

Answer from: Quest

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

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Answer from: Quest

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

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