A coin of mass 25g is pushed on a table. The coin starts moving at speed of 25m/s and stops on 5 sec. Find the frictional force exerted by the table
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Answer:
F = -0.00125N
explanation:
coin of mass (u) = 25g
time (t) = 5sec
speed of coin (v) = 25m/s
v = u + at
0 = 25 + 5a
- 5a = 25
a = 25/5
a = -5
the frictional force (F) = ma
F = 25/1000×(-5)/100
F = -0.00125N
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