A mass
m
=0.2 kg is hung from a string. The other end of the string is attached
to a PAScar of mass
M
=200 grams. The PAScar and hanging mass start from
rest (
v
i
=
0) and the hanging mass descends a distance of 1 m. Using Eq. 4.5
from the manual, calculate the velocity,
v
f
,
of the PAScar. (1.0 point).
200g*9.8=1960
1960=(400g)a
a=1960/400=4.9m/s^2
1=.5*4.9*t^2+0+0
T=0.639s
0.639*4.9=3.13m/s=final velocity
4.
Let’s consider the system in reverse. The PAScar (
M
=200 g) is moving up the
track with an initial velocity (
v
i
) until it stops (
v
f
=0). What must the initial
velocity be in order to lift a hanging mass
m
=0.15 kg on the string a height of
1 m. (1.0 point).
150g*9.8=1470
1470/(350g)=4.2m/s^2
0=4.2*.5*t^2-1
T=0.69s
0.69*4.2=2.90m/s=required initial velocity
5.For the same amount of work done to move the PAScar, if the mass of the PAScar increased, would the final velocity vfincrease, decrease, or stay the same? Explain why using conservation of energy concepts. (2.0 points)

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