Question

a stone of 1 kg is thrown with the velocity of 20 ms-1 across the frozen surface of lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and ice
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Answer 1

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 the opposing force which is Friction)

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Answer 2

Answer:

$\huge\colorbox{yellow}{Given\:-}$

$m = 1 \: kg$

${u}^{2} = 20 \: m \: {s}^{ - 1}$

$s = 50 \: m$

$\huge\colorbox{yellow}{To\:Find\:-}$

Force of friction between the stone and ice.

$\huge\colorbox{yellow}{Solution\:-}$

$\huge{\boxed{\green{ {v}^{2} = {u}^{2} + 2as}}}$

$0 = 400 + 2 \times a \times 50$

$0 = 400 + 100a$

$- 400 = 100a$

$a = \frac{ - 400}{100}$

$a = - 4$

Now, the force of friction

$\huge{\boxed{\red{F = m \times a}}}$

$F = 1 \times - 4$

$\boxed{\pink{F = - 4 \: N}}$

$\huge\colorbox{yellow}{Thank\:You}$