A stone of mass 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake. It 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 mass 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake. It 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
Given, Initial velocity of the stone = 20 m/s.
Final velocity of the stone, v = 0 m/s [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=−4 m/s2.
The negative sign indicates that acceleration is acting against the motion of the stone.
Now, mass of the stone, m = 1 kg.
From Newton's second law of motion:
Force = mass x 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
Given, Initial velocity of the stone = 20 m/s.
Final velocity of the stone, v = 0 m/s [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=−4 m/s2.
The negative sign indicates that acceleration is acting against the motion of the stone.
Now, mass of the stone, m = 1 kg.
From Newton's second law of motion:
Force = mass x 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.
