A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and tie?
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Answered by
1
hey friend , I am having an answer for you ,
m= 1 kg
u = 20m/s
v = 0m/ss(distance travelled) = 50mv²=u²+2as0² = (20)²+2(a)(50)-400 = 100aa = -400/100 = -4m/s² (retardation)F = m×aF = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)hope this helps you friend
thank you !
m= 1 kg
u = 20m/s
v = 0m/ss(distance travelled) = 50mv²=u²+2as0² = (20)²+2(a)(50)-400 = 100aa = -400/100 = -4m/s² (retardation)F = m×aF = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)hope this helps you friend
thank you !
Answered by
0
v= om/s
u=20m/s
s=50m
m= 1kg
to know accerlaration
2as=v^2-u^2
2×a×50= 0-400
100a= -400
a= -400/100
a= -4
to know the force of friction
f=m×a
f= 1×-4
f=-4N
u=20m/s
s=50m
m= 1kg
to know accerlaration
2as=v^2-u^2
2×a×50= 0-400
100a= -400
a= -400/100
a= -4
to know the force of friction
f=m×a
f= 1×-4
f=-4N
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