(1) A 36 kg object is placed on a table. How much force is required to be applied on the object so that is moves with "acceleration" 15 m/s2.
(2) A particle starts from rest and moves with uniform acceleration. It moves a distance of 91 feet in the 7th second.Find its acceleration.
2007-06-25 23:54:14 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
1) 540 N
2) 2.054 m/s^2
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2007-06-26 07:52:12 · answer #1 · answered by Prashant 6 · 0⤊ 0⤋
(1) (36 kg)(15 m/s^2) = 540 N
(2) s = s0 + v0t + (1/2)at^2
s(8) - s(7) = (1/2)a(64 - 49) = 91
a = 182/15 = 5.466667 m/s^2
if t=0 to t=1 is the 0th second
or
a = 182/13 = 14 m/s^2
if t=0 to t=1 is the first second.
Usually, the latter is the way we count.
2007-06-26 14:33:29 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
(1)F=m*a
=36*15
=540N
(2)a=(final velocity-initial velocity)/t
=(91-0)/7
=13ft/s^2
=3.9624m/s^2
2007-06-26 07:39:35 · answer #3 · answered by Shy Lad 3 · 0⤊ 0⤋