Question

a stone of 25 kg is thrown with a velocity of 400 m/s across the ffrozen surface of a lake and comes to rest aafter travelling a distance of 90 m what is the force of friction between the stone and he ice

Answer 1

HELLO FRIEND HERE IS UR ANSWER

Mass= 25 Kilograms

U= 400 m/s

V=0 m/s

Distance= 90 Meters

By using third equation of motion

V2-U2= 2as

0= (400)2+2a(90)

-1600=180a

a= -1600/180= -8.8m/S2

F= 25*8.8= -220

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

u = 400 m/s

v = 0 m/s

s = 90 m

According to the third equation of motion:

v^2 = u^2 + 2as

(0)^2 = (400)^2 + 2 × a × 90

a = – 888.88 m/s2

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

m = 25 kg

From Newton's second law of motion:

F = Mass x Acceleration

F= ma

F= 25× (– 888.88) = – 22222.22 N

Hence, the force of friction between the stone and the ice is – 22222.22 N.

I hope, this will help you.

Thank you______❤

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