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WORK ENERGY AND POWER Quiz-14

Dear Readers,

Work power energy is the most important chapter when it comes to mechanics, for JEE (Advanced). The chapter is quite tricky and takes a lot of time and devotion on your part to understand and master. But this chapter is a complete gold medal for almost all the questions of the mechanics can be solved by the work power energy approach if you master this topic..

Q1.
  •  0 J
  •  +600 J
  •  -600 J
  •  +1600 J
Solution
f=3/10 mg
W=-fs =-3/10 mgS
=-3/10×200×10 J=-600 J


Q2.In the above questions, the work done by frictional force is
  •  +1600 J
  •  -1600 J
  •  +3200 J
  •  -3200 J
Solution
Work done by friction force is equal to negative of the work done by the coolies


Q3.  
  •  Only statement I is true
  •  Only statement II is true
  •  Only I and III are true
  •  Only II and III are true
Solution

Q4. A spring is compressed between two toy carts of masses m_1 andm_2. When the toy carts are released, the spring exerts on each toy cart equal and opposite forces for the same small timet. If the coefficients of friction μ between the ground and the toy carts are equal, then the magnitude of displacements of the toy carts are in the ration
  •  s1/s2 =m2/m1
  •  s1/s2 =m1/m2
  •  s1/s2 =(m2/m1 )^2
  •  s1/s2 =(m1/m2 )^2
Solution
Minimum stopping distance =s
Work done against the friction =W=μmgs
Initial momentum gained by both toy carts will be same
Because same force acts for same time Initial kinetic energy of the toy cart =(p^2/2m) Therefore, (μmgsp^2)/2m or s=(p^2/(2μgm^2 ))
For the two toy carts, momentum is numerically the same. Further μ and g are the same for the toy carts. So s1/s2 =(m2/m1 )^2


Q5.Let r be the distance of a particle from a fixed point to which it is attracted an inverse square law force given by F=k/r^2 (k= constant). Let m be the mass of the particle and L be its angular momentum with respect to the fixed point. Which of the following formulae is correct about the total energy of the system
  •  
  •  
  •  
  •  
Solution
 


Q6. Two trucks A andB, one loaded and the other unloaded, are moving and have same kinetic energy. The mass of A is double that ofB. Brakes (providing same retarding force) are applied to both and are brought to rest. If the distance converted by A before coming to rest is s1 and that of B is s2, then
  •  s1=s2
  •  s1=2s2
  •  2s1=s2
  •  s1=4s2
Solution
Work done by braking force=change in KE -Fs=0-KEi ⇒s=(KEi)/F KEi and F are same for both, Hence, s is also same for both


Q7.A force F=-K(yi+xj) (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a,0) and then parallel to the y-axis to the point (a,a). The total work done by the force F on the particles is
  •  -2 Ka^2
  •  2Ka^2
  •  -Ka^2
  •  Ka^2
Solution


Q8.A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then
  •  Lorry will come to rest in a shorter distance
  •  Car will come to rest in a shorter distance
  •  Both come to rest in a same distance
  •  None of the above
Solution
Stopping distance =(kinetic energy)/(retarding force)⇒s=1/2 (mu^2)/F
If lorry and car both possess same kinetic energy and retarding force is also equal then both come to rest in the same distance


Q9. The energy required to accelerate a car from rest to 10 ms^(-1) is E. What energy will be required to accelerate the car from 10 ms^(-1) to 20 ms^(-1)?
  •  E
  •  3E
  •  5E
  •  7E
Solution


Q10. A particle is projected with a velocity u making an angle θ with the horizontal. The instantaneous power of the gravitational force

  •  Varies linearly with time
  •  Is constant throughout
  •  Is negative for complete path
  • None of these
Solution

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