A 1.40-kg mass is attached to a spring with spring constant 32.0 N/m on a fricti
ID: 1477875 • Letter: A
Question
A 1.40-kg mass is attached to a spring with spring constant 32.0 N/m on a frictionless, horizontal table. The spring–mass system is stretched to 3.50 cm beyond the equilibrium position of the spring and is released from rest at t = 0.
(a) What is the maximum speed of the 1.40-kg mass?
(b) What is the maximum acceleration of the 1.40-kg mass?
(c) What are the position, velocity, and acceleration of the 1.40-kg mass as functions of time? (Use the following as necessary: t. Do not enter units in your expression. Assume t is in seconds.)
cm
x(t) =cm
v(t) = cm/s a(t) = cm/s2Explanation / Answer
given
m = 1.4 kg
k = 32 N/m
A = 3.5 cm = 0.035 m
a) Apply conservation of energy
maximum kinetic enrgy = maximum potentail energy stored in the spring
0.5*m*vmax^2 = 0.5*k*A^2
Vmax^2 = k*A^2/m
Vmax = A*sqrt(k/m)
= 0.035*sqrt(32/1.4)
= 0.167 m/s
= 16.7 cm/s
b) Ac(max)
a = - (0.035)* (32/1.40)
a = -0.8 m/s^2
a= -80 cm/s^2
c)
X=Acos(wt)
X=3.5 cos(4.78t)
V= -Aw sin(4.78t)
V= -16.73 sin(4.78t)
a=-Aw^2 cos (wt)
a= -79.97 cos(4.78t)
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