Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on
1 answer:
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Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)
Answer:
t = 1.09 s.
Explanation:
This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.
This quadratic equation can be solved using determinant.
Of course, we will choose the positive time.