Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring...
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Physics
Inside a NASA test vehicle, a 3.50-kg ball is pulled along by ahorizontal ideal spring fixed to a friction-free table. The forceconstant of the spring is 227 N/m . The vehicle has a steadyacceleration of 5.00 m/s2, and the ball is not oscillating.Suddenly, when the vehicle's speed has reached 45.0 m/s, itsengines turn off, thus eliminating its acceleration but not itsvelocity.
A. Find the amplitude.
B. Find the frequency of the resulting oscillations of the ball.(i already found the answer for this to be 1.28 Hz)
C. What will be the ball's maximum speed relative to thevehicle?
Inside a NASA test vehicle, a 3.50-kg ball is pulled along by ahorizontal ideal spring fixed to a friction-free table. The forceconstant of the spring is 227 N/m . The vehicle has a steadyacceleration of 5.00 m/s2, and the ball is not oscillating.Suddenly, when the vehicle's speed has reached 45.0 m/s, itsengines turn off, thus eliminating its acceleration but not itsvelocity.
A. Find the amplitude.
B. Find the frequency of the resulting oscillations of the ball.(i already found the answer for this to be 1.28 Hz)
C. What will be the ball's maximum speed relative to thevehicle?
Answer & Explanation Solved by verified expert
Find the force on the ball  Â
F = m a
= ( 3.5 kg ) ( 5.00 m/s2)
=Â Â Â 17.50Â Â Â N
( a.)
As the ball is not oscillating, this must be equal to the restoring force exerted by the spring.
   F   =   k x
Hence the maximum displacement (amplitude) of the ball  Â
AÂ Â Â =Â Â Â F / k
=Â Â Â 17.50 / 227
= 0.0771 m
-----------------------------------------------------------------------------------------------------------------------------------
(b) Expression for frequency:
f   =   (1/2π) √(k/m)
=   (1/2 * 3.14) √(227 / 3.5)
=Â Â Â 1.28 Hz
---------------------------------------------------------------------------------------------------------------------------------
( c) Expression for maximum velocity:
v   =   ω A
=   2 π f A
=   2 π ( 1.28 ) ( 0.0771 m)
=Â Â Â 0.621 m/s
   Thus, the maximum velocity relative to the vehicle
   vmax = vvehicle + v
            =   45.0 + 0.621 m/s
            =   45.62 m/s
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