Free falling body's decelerationOk, a body is falling with a velocity 70 2 m s − 1 and its final velocity must be 0 m s − 1 after falling a further 200 m. What is the acceleration required to bring it to a stop after 200 m?Using the formula v 2 = v 0 2 + 2 a xand substituting in the known quantities gives 0 = 9800 + ( 2 a × 200 ) .Rearranging for a gives a = − 9800 400 = − 24.5 m s − 2 . Ok that's easy enough and that's the given answer, but should we not account for g in the answer also?I mean the acceleration due to gravity hasn't vanished. 'It wants' to speed the body up, so shouldn't the deceleration be ( 24.5 + g ) m s − 2 ?

Question

Free falling body&#039;s decelerationOk, a body is falling with a velocity   70      2       m      s          −      1       and its final velocity must be 0   m      s          −      1       after falling a further 200   m. What is the acceleration required to bring it to a stop after   200     m?Using the formula      v    2    =      v    0    2    +  2  a  xand substituting in the known quantities gives  0  =  9800  +  (  2  a  ×  200  )  .Rearranging for   a gives                    a                            =        −                  9800          400                                                                =        −        24.5                 m                  s                      −            2                          .            Ok that&#039;s easy enough and that&#039;s the given answer, but should we not account for   g in the answer also?I mean the acceleration due to gravity hasn&#039;t vanished. &#039;It wants&#039; to speed the body up, so shouldn&#039;t the deceleration be   (  24.5  +  g  )     m      s          −      2      ?

Aiden Villa · Accepted Answer

I think you&#039;re mentally confusing &quot;acceleration&quot; and &quot;force.I think your thought process is &quot;If I&#039;m going to apply an acceleration to an object, that acceleration will be fighting against the acceleration of gravity, so the two accelerations will partially cancel each other out&quot;.The problem is that acceleration isn&#039;t something you can &quot;apply&quot; to an object - you can apply a force. The acceleration is just something you can measure about an object&#039;s motion. So, you&#039;re not &quot;applying&quot; an acceleration of   −  24.5     m      s          −      2      , the object is, according to the problem, accelerating at that rate.If the problem asked you to find the force necessary to accomplish this, then you would indeed need to account for gravity, because the force that would result in such an acceleration would have to be larger since it&#039;s fighting against the influence of gravity:      F          n      e      t        =  m  a  Σ  F  =      F          a      p      p      l      i      e      d        +      F    g    =  m  a      F          a      p      p      l      i      e      d        =  m  a  −      F    g        F          a      p      p      l      i      e      d        =  m  a  −  m  gThe acceleration is merely a measurement of the motion that is described in the problem; there may be many forces on the object including gravity, but we aren&#039;t asked to calculate them.

Abdiel Mays · Accepted Answer

Suppose you apply an upwards force   F on the falling body. Gravity produces a downwards force   −  m  g so the net force is:      F          n      e      t        =  F  −  m  gThe net acceleration is then:      a          n      e      t        =            F              n        e        t              m    =      F    m    −  gThe acceleration you have calculated is the net acceleration       a          n      e      t       given by the equation above.

Answered question

Answer & Explanation

New Questions in Force, Motion and Energy