Question

Mass of 1kg is thrown with velocity 20 ms and comes back to rest after travelling 50 metres . What is the force of friction between stone and ice

Answer 1

Answer:

M = 1kg

u = 20m/s. v = 0m/s.

s(distance travelled) = 50m

using third equation of motion

v²=u²+2as

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

F = m×a

F = 1×(-4) = -4N.

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Answer 2

Answer:

m = 1kg

u = 20m/s. v = 0m/s.

s(distance travelled) = 50m

using third equation of motion

v²=u²+2as

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

F = m×a

F = 1×(-4) = -4N. (negative sign indicates)

#Hope it helps✌