Work Energy and Power-Quiz-21

WORK ENERGY AND POWER Quiz-21

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.In a conservative force field, we can find the radial component of force F from the potential energy function (U) using the relation F=dU/dr. Positive value of F mean repulsive forces and vice-versa. We can find the equilibrium position, where force is zero. We can also calculate ionisation energy, which is the work done to move the particle from a certain position to infinity. Let us consider a particle bound to a certain point at a distance r from the centre of the force. The potential energy function of the particle is given by U (r)=A/r^2 -B/r, where A and B are positive constants 

 The nature of equilibrium is

•  Neutral
•  Stable
•  Unstable
•  Cannot the predicted

Solution

Q2.Two unequal masses are tied together with a cord with a compressed spring in between When the cord is burnt with a match releasing the spring, the two masses fly apart with equal

•  Kinetic energy
•  Speed
•  Momentum
•  Acceleration

Solution

Q3.  A body of mass 2 kg starts from rest and moves with uniform acceleration. It acquires a velocity 20 ms^(-1) in 4 s The power exerted on the body at 2 s is

•   50 W
•  100 W
•  150 W
•  200 W

Solution
v=u+at ⇒20=0+a×4 ⇒a=5 ms^(-2)
F=ma=2×5=10 N
Velocity at 2 s:v_1=at=5×2=10 ms^(-1)
Power =Fv_1=10×10=100 W

Q4. 

Sand particles drop vertically at the rate of 2 kgs^(-1) on a conveyor belt moving horizontally with a velocity of 0.2 ms^(-1) The extra force required to keep the belt moving is

•  0.4 N
•  0.08 N
•  0.04 N
•  0.2 N

Solution

Q5.A ladder of length l carrying a man of mass m at its end is attached to the basket of a balloon of massM. The entire system is in equilibrium in the air. As the man climbs up the ladder into the balloon, the balloon descends by a height h
The potential energy of the man

•  Increases by mg(l-h)
•  Increases by mgl
•  Increases by mgh
•  Increases by mg(2l-h)

Solution
 When the man climbs the ladder of length l, the balloon descends by height h. So net height gained by the man is l-h. Hence, gain in potential energy of the man is mg(l-h) Work done by man :W_1=mgl Inclination in potential energy of balloon= work done by man – increase in potential energy of man, i.e., mgl-mg(l-h)=mgh

Q6. 

A single conservation force F(x) acts on a 1.0 kg particle that moves along the x-axis. The potential energy U(x) is given by U(x)=20+(x-2)^2 where x is in meters. At x=5.0 m, the particle has a kinetic energy of 20 J What is the mechanical energy of the system?

•  35 J
•  64 J
• 86 J
•  49 J

Solution

At x=5 m, U=20+(5-2)^2=29 J, K=20 J Mechanical energy=E=U+K=29+20=49 J

Q7.A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by F ⃗=(4-x^2)i ̂ N, when x is in meter and the initial position of the block is x=0 The maximum kinetic energy of the block between x=0 and x=2.0 m is

•  2.33 J
•  8.67 J
•  5.33 J
•  6.67 J

Solution

Q8.

•  F(2 R)
•  TF(√2 R)
•  πF(√2 R)
•  F (2R)

Solution

Q9.

•  
  
•  
  
•  
  
•  
  

Solution

Loss in PE of spring + loss in PE of m_2= gain in KE of (m_1+m_2 )+work done against friction ⇒

 1/2 kx^2+m_2 gh=1/2 (m_1+m_2 ) v^2+μ_k m_1 gh ⇒

 v=√((kx^2+2m_2 gh-2μ_k m_1 gh)/(m_1+m_2 ))

Q10. 

•  9.2 ms^(-1)
•  9 ms^(-1)
•  6.5 ms^(-1)
• 8 ms^(-1)

Solution