Science, asked by chadrackbchandra, 6 months ago

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

Answers

Answered by lgrajan2002
12

Explanation:

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)

hope this helps

Answered by stuanumeet2425
0

Answer: Given: m = 1kg , u = 20m/s , v = 0m/s, s = 50 m

to find: acceleration(a) = ?  and force of friction(f)

solution : to find acceleration(a)

we will use the kinametical equation of motion i.e. 2as = final velocity x final velocity - initial velocity x  initial velocity

we can write it as : a = v2 - u2\2s

 =  0 * 0 - 20 * 20/2(50)

 = 0 - 400/100

 = -400/100

 = -4/1 = -4m/s2

 a = -4m/s2

now to find force we will use the formula f = ma

f = 1 * (-4)

f =  -4N

 

Explanation:

the final velocity is 0 because it is given that the stone comes to rest .

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