A block is released from rest and falls vertically with a vertical spring attach
ID: 1717361 • Letter: A
Question
A block is released from rest and falls vertically with a vertical spring attached. The acceleration of the block is a = (32.2-50s) ft/s^2 where s is measured in feet from the release point positive down. Determine the maximum distance the mass moves down and the maximum velocity of the mass during this downward motion. A steel ball is released from rest in a container of oil. Its downward acceleration is a = 2.4 - 0.6 v in/s^2, where v is the ball's velocity in in/s. What is the ball's downward velocity 2 s after it is released?Explanation / Answer
>>As, acceleration, a = 2.4 - 0.6*v in/s2 ,
where, v = velovity of ball (in/s)
>> Initially, ball is at rest, => at t = 0, v = 0
>> As, a = dv/dt
=> dv = a*dt
>> (2.4 - 0.6*v) dt = dv
>> Integrating both sides from t= 0 to 't' seconds
=> (2.4*t - 0.3*t2 )0t= |v|0V(t)
=> V(t) = 2.4t - 0.3*t2 ........Velocity as function of 't'........ANSWER....
>> At t = 2 sec,
=> V(2) = 2.4*2 - 0.3*2*2 = 3.6 in/s
=> Velocity at time 2sec = 3.6 in/s ............ANSWER.........
>> As, v = ds/dt
=> v*dt =ds
>> (2.4*t - 0.3*t2) dt = ds
>> Integrating both sides from t= 0 to 't' seconds
=> (1.2*t2 - 0.1*t3 )0t= |s|0s(t)
=> s(t) = 1.2*t2 - 0.1*t3 .................Displacement as function of time,'t' ........ANSWER.......
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