## WORK ENERGY AND POWER Quiz-6

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.
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Solution
Solve this question relative to the frame (car) of reference. For maximum velocity (relative to frame), the block must be in equilibrium position Let x_0 be the equilibrium elongation in spring, then

Q2.
A toy gun uses a spring of force constantK. Before being triggered in the upward direction, the spring is compressed by a distancex. If the mass of the shot is m, on being triggered it will go up to a maximum height of
•  (Kx^2)/mg
•  x^2/Kmg
•  (Kx^2)/2mg
•  (K^2 x^2)/mg
Solution

Q3.
•  10 J
•  100 J
•  0.01 J
•  1 J
Solution

Q4. A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line asv=k √x, where k is a positive constant. The work done by all the forces acting on the particle, during the first t seconds is
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Solution

Q5.
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Solution

Q6.
•  If it is released at the origin, it will move in negative x-axis
•  If it is released at x=2+Î” where Î”→0, then its maximum speed will be 5 ms^(-1) and it will perform oscillatory motion
• If initially x=-10 and u =√6 i ̂, then it will cross x=10
•  x=-5 and x=+5 are unstable equilibrium positions of the particle
Solution

Q7.
•  same
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•  None of the above
Solution

Q8.
•  360 N
•  720 N
•  1440 N
•  2880 N
Solution

Q9. An engine pumps up 100 kg of water through a height of 10 m in 5 s. given that the efficiency of the engine is 60 %. What is the power of the engine? Take g=10 ms^(-2)
•  33 kW
•  3.3 kW
•  0.33 kW
•  0.033 kW
Solution
Power used to pump the water =mgh/t=(100×10×10)/5 =200 W Power of engine 4 =2000×100/60=3.3 kW

Q10.

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250 m/s
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250√2 m/s
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400 m/s
• 500 m/s
Solution
R=u√(2h/g)⇒20=V1 √((2×5)/10) and 100=V2 √((2×5)/10) ⇒V1=20 m/s,V2=100 m/s
Applying momentum conservation just before and just after the collision (0.01)(V)=(0.2)(20)+(0.01)(100)
V=500 m/s

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Work Energy and Power-Quiz-6