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
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
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 .