A stone of 1 kg is thrown with a velocity of 20 ms−1 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 the ice?
A stone of 1 kg is thrown with a velocity of 20 ms−1 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 the ice?
The correct option is D 4N
Initial velocity of the stone = 20 ms−1 .
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=20×20+2×a×50
a=−4 ms−2.
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) = - 4 N
Hence, the force of friction between the stone and the ice is 4 N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.
4N
Initial velocity of the stone = 20 ms−1 .
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=20×20+2×a×50
a=−4 ms−2.
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) = - 4 N
Hence, the force of friction between the stone and the ice is 4 N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.
