Work Energy and Power-Quiz-13

WORK ENERGY AND POWER Quiz-13

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.

Two particles A and B are interacting with each other only. Initially, particle A has a speed of 5 m/s and particle B has a speed of 10 m/s. In the course of their motion, they return to their initial position. Finally A has a speed of 4 m/s and B ha speed of 7 m/s. We can conclude that

•  The potential energy changed from beginning to the end of trip
•  The mechanical energy was increased by non-conservative forces
•  The mechanical energy was decreased by non-conservative forces<
•  None of these

Solution

From the data, it is clear that mechanical energy is decreased

Q2.

•  2.4 J
•  5.55 J
•  4.44 J
•  10 J

Solution

Q3.  

•  
  
•  
  
•  
  
•  
  

Solution

Q4. 

An electric motor creates a tension of 4500 N in hoisting a cable and reels it at rate of 2 ms^(-1). The power of the electric motor is

•  15 kW
•  9 kW
•  225 W
•  9000 H.P.

Solution

P=Tv (numerically)
=4500×2=9000 W=6 kW

Q5.

•  4 ms^(-1)
•  2 ms^(-1)
•  1 ms^(-1)
•  3 ms^(-1)

Solution
 

Q6. 

• w1>w2>w3
• w1=w2=w3
• w1<w2<w3
• w2>w1>w3

Solution

Gravitational field is a conservative force field. In a conservative force field work done is path independent. ∴ W1=W2=W3

Q7.

•  Zero
•  √(2E/m)
•  √(E/m)
•  √(3/2m)

Solution

From the conservation of energy
KE+PE=E or KE=E-1/2 kx^2
KE at x=-√(2E/k) is E-1/2 k(2E/k)=0
The speed of particle at x=-√(2E/k) is zero

Q8.

In the above question, the average power delivered by gravity is

•  -mg u cos⁡α
•  -mgu sin⁡α
•  -(mgu cos⁡α)/2
•  -(mgu sin⁡α)/2

Solution

Q9. 

•  
  
•  
  
•  
  
•  
  

Solution

Q10. 

A 500 kg car, moving with a velocity of 36 kmh^(-1) on a straight road unidirectionally, doubles its velocity in 1 min. The average power delivered by the engine for doubling the velocity is

•  750 W
•  1050 W
•  1150 W
• 1250 W

Solution

u=10 ms^(-1),v=20 ms^(-1)
Work done = Increase in kinetic energy
=1/2×500[20^2-10^2 ]=(500×30×10)/2
Power=(500×30×20)/(2×60)=1250 W