It results change only in it's kinetic energy, it's KE will increase in accord with the work-energy theorem
Explanation:
Show that the motion of a mass attached to the end of a spring is SHM
Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed
at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.
If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.
According to "Hook's Law
F = - Kx ---- (1)
Negative sign indicates that the elastic restoring force is opposite to the displacement.
Where K= Spring Constant
If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion
from a to b and then b to a.
According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by
F = ma ---- (2)
Comparing equation (1) & (2)
ma = -kx
Here k/m is constant term, therefore ,
a = - (Constant)x
or
a a -x
This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.
Answer: 900
Explanation: Force equals mass x acceleration F=M×A
Answer:
W = - 118.24 J (negative sign shows that work is done on piston)
Explanation:
First, we find the change in internal energy of the diatomic gas by using the following formula:
where,
ΔU = Change in internal energy of gas = ?
n = no. of moles of gas = 0.0884 mole
Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)
Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K
ΔT = Rise in Temperature = 18.8 K
Therefore,
Now, we can apply First Law of Thermodynamics as follows:
where,
ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)
W = Work done = ?
Therefore,
<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>
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