A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and ice.
Please reply fast
Answers
Answered by
8
deceleration of stone = a
v² - u² = 2 a s
0² - 20² = 2 a * 50 => a = - 4 m/s²
force = m a = - 4 * 1 = - 4 N
v² - u² = 2 a s
0² - 20² = 2 a * 50 => a = - 4 m/s²
force = m a = - 4 * 1 = - 4 N
shanmu54321:
tnx a ton
Answered by
15
initial velocity (u)= 20 m/s
final velocity (v) = 0 m/s
distance covered (s) = 50 m
mass (m) = 1 kg
Acc. to third eq. of motion
v2 = u2 + 2as
so on solving it
we get acceleration (a) = 4m/s2
so, force of friction (f) = 4 N
final velocity (v) = 0 m/s
distance covered (s) = 50 m
mass (m) = 1 kg
Acc. to third eq. of motion
v2 = u2 + 2as
so on solving it
we get acceleration (a) = 4m/s2
so, force of friction (f) = 4 N
Similar questions