Answer :
2nd law: F = ma
hook's law: F = -kx
[tex]ma = -kx[/tex]
[tex]a = \frac{d^2x}{dt^2} = - \frac{k}{m} x [/tex]
try solution: [tex]x = e^{ \lambda t} , \frac{dx}{dt} = \lambda e^{ \lambda t}, \frac{d^2x}{dt^2} = \lambda^2 e^{ \lambda t}[/tex]
[tex] \lambda^2e^{ \lambda t}= - \frac{k}{m}e^{ \lambda t}[/tex]
divide by: [tex]e^{ \lambda t}[/tex]
[tex] \lambda^2 = - \frac{k}{m} [/tex]
solution:
[tex]a =-\frac{k}{m}e^{i \sqrt{ \frac{k}{m} }t } [/tex]
hook's law: F = -kx
[tex]ma = -kx[/tex]
[tex]a = \frac{d^2x}{dt^2} = - \frac{k}{m} x [/tex]
try solution: [tex]x = e^{ \lambda t} , \frac{dx}{dt} = \lambda e^{ \lambda t}, \frac{d^2x}{dt^2} = \lambda^2 e^{ \lambda t}[/tex]
[tex] \lambda^2e^{ \lambda t}= - \frac{k}{m}e^{ \lambda t}[/tex]
divide by: [tex]e^{ \lambda t}[/tex]
[tex] \lambda^2 = - \frac{k}{m} [/tex]
solution:
[tex]a =-\frac{k}{m}e^{i \sqrt{ \frac{k}{m} }t } [/tex]
The combined formula of Newton's second law and Hook's law to determine acceleration of a block as function of time is [tex]a(t) = \frac{k.t(v_0 + v_f)}{2m}[/tex]
According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.
F = ma
The Hook's law is given as;
F = kx
Combining the two equations together;
ma = kx
[tex]a = \frac{kx}{m}[/tex]
where;
x is the displacement or extension of the spring
[tex]x = (\frac{v_0 + v_f}{2} )t\\\\a = \frac{kx}{m} = \frac{k((\frac{v_0 + v_f}{2} )t)}{m}\\\\a = \frac{kt(v_0 + v_f)}{2m}[/tex]
Thus, the combined formula of Newton's second law and Hook's law to determine acceleration of a block as function of time is [tex]a(t) = \frac{k.t(v_0 + v_f)}{2m}[/tex]
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