A coin of mass 25g is pushed on a table. The coin starts moving at speed of 25cm/s and stops in 5s. Find the force of friction exerted by the table on the coin.
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t= 5s
initial speed, u = 25cm/s
final speed,v = 0 cm/s
the coin will come to rest because of the retarding nature of the frictional force.
if s, is the distance travelled by the coin, then
s = ut +at²/2
v²=u²+2as
v²=u²+2a(ut+at²/2)
v²=u²+2aut+a²t²
0=25²+2×a×25×5+a²×25²
0=25+2×5×a +25a²
0= (5a+1)²
a= -1/5 cm/s²
, the minus sign means that the frictional frce was applied in the direction opposite to the direction of motion.
initial speed, u = 25cm/s
final speed,v = 0 cm/s
the coin will come to rest because of the retarding nature of the frictional force.
if s, is the distance travelled by the coin, then
s = ut +at²/2
v²=u²+2as
v²=u²+2a(ut+at²/2)
v²=u²+2aut+a²t²
0=25²+2×a×25×5+a²×25²
0=25+2×5×a +25a²
0= (5a+1)²
a= -1/5 cm/s²
, the minus sign means that the frictional frce was applied in the direction opposite to the direction of motion.
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