## WORK ENERGY AND POWER Quiz-15

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.
•  U=-2x^2 y
•  U=-2x^2 y+ constant
•  U=2x^2 y+ constant
•  No defined
Solution

Q2.
•  25 cm
•  37.5 cm
•  62.5 cm
•  None of the above
Solution
W_gravity=∆U_spring
mg sin⁡Î¸ (d+0.25)=1/2 k(0.25)^2
Which gives d=37.5 cm

Q3.
•  4
•  2
•  1/2
•  1/4
Solution
From energy conservation, 1/2 kx^2=1/2 (4k) y^2
y/x=1/2

Q4. When a person stands on a weighing balance, working on the principle of Hooke’s law, it shows a reading of 60 kg after a long time and the spring gets compressed by 2.5 cm. If the person jumps on the balance from a height of 10 cm, the maximum reading of the balance will be
•  60 kg
•  120 kg
•  180 kg
•  240 kg
Solution

Q5.
•
•
•
•
Solution

Q6. A person of mass 70 kg jumps from a stationary helicopter with the parachute open. As he falls through 50 m height, he gains a speed of 20 ms^(-1). The work done by the viscous air drag is
•  21000 J
•  -21000 J
•  -14000 J
•  14000 J
Solution
From work-energy theorem, net work done by all forces (internal and external)=change in kinetic energy
⇒ Work done by gravity + work done by air drag
=1/2×70(20^2-0^2 )
⇒ Work done by air drag=14000-35000=-21000 J

Q7.

•  4.50 J
•  7.50 J
•  5.06 J
•  14.06 J
Solution

Q8.

•  -10 J
•  100 J
•  10 J
•  1 J
Solution

Q9. A particle of mass m moves along a circular path a radius r with a centripetal acceleration a_n changing with time t as an=kt^2, where k is a positive constant. The average power developed by all the forces acting on the particle during the first t_0 seconds is
•  mkrt0
•  (mkrt0^2)/2
•  (mkrt0)/3
•  (mkrt0)/4
Solution

Q10.

•  8 J
•  9 J
•  7 J
• 0.48 J
Solution

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