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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Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions
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(b) 0.5 mole of water molecules?
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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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(d) What is the effect of change of temperature on the solubility of a salt?
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