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Problems on Work
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This lesson deals with the problems involving the concepts discussed in the previous lessons.

Adarsh Raj
Teaching Physics for IIT JEE /NEET. 20 MCQ for NEET daily In my feed. Question Series On Each chapter. HC VERMA EXERCISE solving .

U
Unacademy user
Sir what is the answer with slipping?
Rishabh Sharma
2 years ago
In that case, Point of Contact will have acceleration : vector addition of (centripetal + Linear acceleration term - angular acceleration term).
Jasvinder Singh
2 years ago
Thq sir
  1. Adarsh Raj E.C.E Educator at Unacademy Loves teaching and debating https://unacademy.com/user/AdarshRajSrivastav.


  2. Q-A particle of mass 2 kg moves under the action of a constant force F = (51-2) N. lf its displacement is 6i m, what is the work done by the force F 2 The force acting on the body is E = (5-2 j )N while the displacement, The work done = FX = (5-2 j )6: -30 joule Q-A tug, exerting a pulling force of 800 N due north, tows a barge through a distance of 1 km in a direction 3d E of north. What is the work done by the tug ?


  3. Problem A particle of mass m is displaced from a position Pi to P2 with position vectors F1 = ai + bj + ck and ,-ci + aj-bk by a force F = b i + c j + ak . Find the work done by the force. If the displacement of the particle is s , the work done W by the given force F is equal to F .s , when F is a constant force Solutionn where s-Ar = r,-r, s =[(citaj + b k )-(a i+bj + ck )] S =[( c-aji + (a-b))+(b-c)k] and F bi cjak Using (1), (2) and (3) w = (bi +cj +ak).[( c-aji +(a-b)) +(b-c)k] w = (c-a)b + (a-b)c + (b-c)a W=bc-ab + ac-bc + ab-ac=0 Net work done by the force F for the given displacement is zero


  4. A particle is taken from point P to point Q via the path PAQ and then placed back to point P via the path QBP. Find the work done by gravity on the body over this closed path. Vertical separation between P and Q is h as shown Here, displacement of the particle is PQ, gravity is acting vertically downward. The vertical component of PQ is h (say) upward. Hence (PAQ) -mgh For the path QBP, component of the displacement along vertical is h(downward) In this case, WQBP) = mgh Total work done = WPAQ + WQBP = 0


  5. Find the work done by the frictional force in drawing a circle of radius r by a pencil of negligible mass with a normal pressing force N (coefficient of friction The kinetic frictional force at any instantaneous position is f.- .N t, acts tangentially opposite to the motion of the pencil. For acts tangentially opposite t elementary displacement d the work done by f k is elementary displacement di the work done by fk is 2 Ttr 2 Ttr


  6. When a man walks on a horizontal surface with constant velocity work done by (A) Friction is zero (C) Gravity is zero B) Contact force is zero (D) Man is zero Since mg & N are perpendicular to velocityv and ds, work done by these forces is zero. Since no relative sliding occurs during walking, static friction comes into play Hence the point of application of static frictional force does not move. Therefore no work is done by frictional force. The man has to lose his body's (internal) energy E, hence performs work because W-AE. d s mg


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