Question

a stone of 1 kg throw with a velocity of 20 metre per second across the frozen surface of a lake and come to rest after traveling the distance of 50 m. what is the force of friction between the stone and ice. give me solutions fast

Answer 1

m = 1kg u = 20m/s. v = 0m/s.  distance travelled = 50m using equation of motion v²=u²+2as 0= (20)²+2(a)(50) -400 = 100a a = -400/100 = -4m/s² (retardation) F = m×a F = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

u = 20 m/s

v = 0 m/s

s = 50 m

According to the third equation of motion:

v^2 = u^2 + 2as

(0)^2 = (20)^2 + 2 × a × 50

a = – 4 m/s2

★The negative sign indicates that acceleration is acting against the motion of the stone.

m = 1 kg

From Newton's second law of motion:

F = Mass x Acceleration

F= ma

F= 1 × (– 4) = – 4 N

Hence, the force of friction between the stone and the ice is – 4 N.

I hope, this will help you.

Thank you______❤

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