A stone of 1 kg is thrown with a velocity of 20 m s-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?
Mass of the stone = 1 kg
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance travelled by the stone, s = 50 m
Using the third equation of motion:
v2 = u2 + 2as
Where,
Acceleration, a
(0)2 = (20)2 + 2 × a × 50
a = −4 m/s2
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
The negative sign shows that force of friction is in the opposite direction of motion , the force of friction between the stone and the ice is −4 N.
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Substance Dissolved | Tem | per | atu | re | in K |
283 | 293 | 313 | 333 | 353 | |
Potassium nitrate | 21 | 32 | 62 | 106 | 167 |
Sodium chloride | 36 | 36 | 36 | 37 | 37 |
Potassium chloride | 35 | 35 | 40 | 46 | 54 |
Ammonium chloride | 24 | 37 | 41 | 55 | 66 |
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