A stone of 1 kg is thrown with a velocity of 20m/s across a frozen surface of a lake and comes to rest after travelling 50 m. What is the force of friction between the stone and the ice?
A stone of 1 kg is thrown with a velocity of 20m/s across a frozen surface of a lake and comes to rest after travelling 50 m. What is the force of friction between the stone and the ice?
Initial velocity of the stone = 20 m/s.
Final velocity of the stone, v = 0 (finally the stone comes to rest).
Distance covered by the stone = 50 m.
According to third equation of motion:
v2=u2+2as
0 × 0 = 20 × 20 + 2 × a × 50
a=−4m/s2.
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg.
From newton's second law of motion:
Force = mass × acceleration. Therefore,
F = 1 × (-4) = - 4N
Hence, the force of friction between the stone and the ice is 4N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.
Initial velocity of the stone = 20 m/s.
Final velocity of the stone, v = 0 (finally the stone comes to rest).
Distance covered by the stone = 50 m.
According to third equation of motion:
v2=u2+2as
0 × 0 = 20 × 20 + 2 × a × 50
a=−4m/s2.
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg.
From newton's second law of motion:
Force = mass × acceleration. Therefore,
F = 1 × (-4) = - 4N
Hence, the force of friction between the stone and the ice is 4N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.
