A 0.50 kg block is attached to a spring of spring constant 32 N/m and set into m
ID: 1981568 • Letter: A
Question
A 0.50 kg block is attached to a spring of spring constant 32 N/m and set into motion. The block oscillates on a horizontal frictionless surface. The amplitude of the motion is 0.10 m.A). What is the magnitude of the block's maximum acceleration?
B). What is the maximum speed of the block?
C). What is the total mechanical energy of the system?
D). When the block is 0.05 m from the equilibrium position, its speed is...
less than half its maximum speed?
exactly half its maximum speed?
more than half its maximum speed?
E). At one point at which the block is instantaneously at rest, you carefully drop an additional mass on top of the block - the block then continues to oscillate, just with extra mass. Increasing the mass like this changes which of the answers to parts A-C above? Select all that apply.
_____The answer to A would now be different
_____The answer to B would now be different
_____The answer to C would now be different
Explanation / Answer
a) max. acceleration means maximum force ,
F= kx
F is maximum at extreme posituions.
F = 32 x 0.10 = 3.2 N
a = F/m = 3.2 / 0.50 = 6.4 m/s2
b) max. speed will be at equilibrium point,
mv2 /2 = kx2 /2
0.50 x v2 = 32 X 0.102
v = 0.8 m/s
c) total mechanical energy of sysytem = kx2 /2 = 32 x 0.12 /2 = 0.16 J
d) 0.5 X v2 /2 = 32 x (0.12 = 0.052 ) /2
v = 0.693 m/s
that greaterr tthan its half of its max. speed (0.8/2) .
e) part A : x is not dpending on mass. so it will same .
part B : yes, max. speed will now decrease. as it depends on mass.
pert c : no , it will not change , it will be equal to kx2 /2 = 0.16 J
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