Complete Question
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?
Answer:
The kinetic energy is ![KE = 0.4368\ J](https://tex.z-dn.net/?f=KE%20%3D%200.4368%5C%20%20J)
Explanation:
From the question we are told that
The mass of the block is ![m= 0.025\ kg](https://tex.z-dn.net/?f=m%3D%200.025%5C%20kg)
The spring constant is ![k = 150 N/m](https://tex.z-dn.net/?f=k%20%3D%20150%20N%2Fm)
The length of first displacement is ![x_1 = 0.80 \ m](https://tex.z-dn.net/?f=x_1%20%3D%200.80%20%5C%20m)
The length of first displacement is ![x_2 = 0.024 \ m](https://tex.z-dn.net/?f=x_2%20%3D%200.024%20%5C%20m)
At the
the kinetic energy is mathematically evaluated as
![KE = \Delta E](https://tex.z-dn.net/?f=KE%20%20%3D%20%5CDelta%20E)
Where
is the change in energy stored on the spring which is mathematically represented as
![\Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20%28x_1%20%5E2%20-%20x_2%5E2%29)
=> ![KE = \frac{1}{2} k (x_1 ^2 - x_2^2)](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20k%20%28x_1%20%5E2%20-%20x_2%5E2%29)
Substituting value
![KE = \frac{1}{2} * 150 * (0.08^2 - 0.024^2)](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20150%20%2A%20%20%280.08%5E2%20-%200.024%5E2%29)
![KE = 0.4368\ J](https://tex.z-dn.net/?f=KE%20%3D%200.4368%5C%20%20J)
A. directly convert hydrogen to electricity
Answer:
know the truth.’ Then I said: ‘Blessed be God, who put this in the Chan’s heart. But our Scriptures tell us, the servant of God should not dispute, but should show mildness
The kinetic energy of a 204 kg motorcycle moves with a velocity of 6 m/s is 3672 joules (3672=0.5*204*6^2). This problem can be solved using the formula of an object's kinetic energy which is the energy of a motion. The formula of an object's kinetic energy is stated as KE = 0.5*m*v^2 where KE is the objects' kinetic energy amount, m is the object's mass, and v is the object's velocity<span>.</span>
We have
y=78.4 m and the distance, x=2400 m
we need to find out the time in the air, t=√2y/a
where a is 9.8 m/s²
t=√ 2*(-78.4)/-9.8 m/s²= √ -156.80/ -9.8 =√16=4s
speed of the bullet will be
v=x/t= 2400/4= 600m/s