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4 0

3 years ago

A 4.00-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 27.0 N is require

NikAS [45]

Answer:

a)

$135$ Nm⁻¹

b)

$0.925$ Hz

c)

$1.2$ ms⁻¹

d)

$0$ m

e)

$6.7$ ms⁻²

f)

$\pm 0.2$ m

Explanation:

a)

$F$ = force required to hold the object at rest connected with stretched spring = 27 N

$x$ = stretch in the spring from equilibrium position = 0.2 m

$k$ = force constant of the spring

force required to hold the object at rest is same as the spring force , hence

$F = k x$

$k = \frac{F}{x}$

inserting the values

$k = \frac{27}{0.2} = 135$ Nm⁻¹

b)

frequency of the oscillations is given as

$f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

inserting the values

$f =\frac{1}{2(3.14) }\sqrt{\frac{135}{4}}\\f = 0.925$ Hz

c)

$A$ = Amplitude of oscillations = 0.2 m

$w$ = angular frequency

Angular frequency is given as

$w = 2\pi f = 2 (3.14) (0.925) = 5.8$ rads⁻¹

Maximum speed of oscillation is given as

$v_{max} = Aw$

$v_{max} = (0.2)(5.8)\\v_{max} = 1.2$ ms⁻¹

d)

maximum speed of the object occurs at the equilibrium position, hence

$x = 0$ m

e)

Maximum acceleration of oscillation is given as

$a_{max} = Aw^{2}$

$a_{max} = (0.2)(5.8)^{2}\\a_{max} = 6.7$ ms⁻²

f)

maximum acceleration occurs when the object is at extreme positions, hence

$x = \pm 0.2$ m

4 0

3 years ago