Work Energy and Power-Quiz-1

WORK ENERGY AND POWER Quiz-1

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.A body is acted upon by a force which is inversely proportional to the distance converted. The work done will be proportional to

•  s
•  s^2
•  √s
•  None of these

Solution

Q2.A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of4 ms^(-1). The tension in the string is 6 N when the stone is

•  At the top of the circle
•  At the bottom of the circle
•  Half way down
•  None of above

Solution

Q3.  A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms^(-1). Which of the following statements(s) is/are correct for the system of these two masses?

•   Total momentum of the system is 3 kg-ms^(-1)
•  Momentum of 5 kg mass after collision is 4 kg-ms^(-1)
•  Kinetic energy of the centre of mass is 0.75 j
•  Total kinetic energy of the system is 4 j

Solution
As the speed of mass is uniform hence, net power will be zero.

Q4. 

•  140 J
•  180 J
•  120 J
•  280 J

Solution
Decrease in height :h=14-7=7 m Also KE at C= Loss in PE=mgh=140 J

Q5.A particle is released one by one from the top of two inclined rough surfaces of height h each. The angles of inclination of the two planes are 30° and60°, respectively. All other factors (e.g., coefficient of friction, mass of block, etc.) are same in both the cases. Let K1 and K2 be the kinetic energies of the particle at the bottom of the plane in the two cases. Then

•  K1=K2
•  K1>K2
•  K1< K2
•  Data insufficient

Solution
 Work done by friction: W=(μmg cos⁡θ )S =(μmg cos⁡θ ) h/sin⁡θ =μmgh cot⁡θ Now cot⁡〖θ1 〗=cot⁡〖30°〗=√3 cot⁡〖θ2 〗=cot⁡〖60°〗=1/√3 i.e., kinetic energy (KE=mgh-W) in first case will be less

Q6. 

•  2/√3
•  √3/2
• √3/4
•  √3/5

Solution

Q7.Two identical particles, A andB, are attached to a string of length 2l,A to middle and B to one of the ends. The string is whirled in a horizontal circle, with the end O fixed. If the kinetic energy of B relative to A is E, then the absolute kinetic energies of A and B are

•  E and E
•  E and 4E
•   4E and E
•  E and 3E

Solution
From v=ωr, speed of B will be double of A, hence KE of B is 4 times that of A

Q8.If a number of forces act on a body and the body is in static or dynamic equilibrium, then

•  Work done by any individual force must be zero
•  TNet work done by all the forces is positive
•  Net work done by all the forces is negative
•  Net work done by all the forces is zero

Solution
When a number of forces act on a body and the body is in static or dynamic equilibrium (it means F_net=0), then individual forces may do the work, but net work done by all the forces will be zero, because KE of body is not changing

Q9.The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as K=as^2. The force acting on the particle is

•  2a s^2/R
•  2as [1+s^2/R^2 ]^(1/2)
•  2as
•  2a

Solution

Q10. A person lifts a bucket full of water by means of light rope from a well. Select the incorrect statement

•  Work done by gravity is negative
•  Work done by the tension in the rope is positive
•  Work done on the person is positive
• Work done on the person is negative

Solution
Here work done by the person is positive, because energy of the person will decrease in lifting the bucket. It means work done on the person should be negative. Also if work done on the person were positive, then his energy should have been increased.