The blocks A and B shown in the figure have masses MA = 5kg and MB = 4 kg. The system is released from rest. The speed of B after A has traveled a distance 1 m along the incline is :
If A moves down the incline by 1 metre. B shall move up by 1/2 metre. If the speed of B is v then the speed of A will be 2v.
Gain in K.E. = loss in P.E.
Solving, we get
The correct answer is:
In the figure the variation of components of acceleration of a particle of mass 1kg is shown w.r.t. time. The initial velocity of the particle is The total work done by the resultant force on the particle in time interval from t = 0 to t = 4 second is :
Your answer is incorrect.
∴ W = 10 J
The correct answer is: 10J
A block of mass m is being pulled up the rough incline by the agent delivering constant power P. The coefficient of friction between the block and the incline is µ. The maximum speed of the block during the courses of ascent is :
Let at any time the speed of the block along the incline upwards be v.
Then from Newton’s second law
the speed is maximum when
The potential energy (in SI units) of a particle of mass 2kg in a conservative fields U = 6x – 8y. If the initial velocity of the particle is then the total distance traveled by the particle in first two seconds is :
has same direction as that of
Since are in same direction, particle will move along a straight line
The correct answer is: 15m
The potential energy of a particle of mass m free to move along x-axis is given by for x < 0 and U = 0 for x ≥ 0 (x denotes the x-coordinate of the particle and k is a positive constant). If the total mechanical energy of the particle is E, then its speed at x = -(2E/k)1/2
From conservation of energy
K.E. + P.E. = E
∴ The speed of particle at is zero.
Work done by static friction on an object :
Consider the blocks shown in the figure to be moving together due to friction between them. The free body diagrams of both the blocks are shown below.
Work done by static friction on A is positive and on B is negative.
The correct answer is: may be positive
A man places a chain (of mass m and length l) on a table slowly. Initially the lower end of the chain just touches the table. The man drops the chain. When half of the chain is in vertical position, the potential energy of chain is :
The work done by man is negative of magnitude of decrease in potential energy of chain
The correct answer is:
A collar B of mass 2kg is constrained to move along a horizontal smooth and fixed circular track of radius 5m. The spring lying in the plane of the circular track and having spring constant 200N/m is undeformed when the collar is at A. If the collar starts from rest at B the normal reaction exerted by the track on the collar when it passes through A is :
Initial extension will be equal to 6 m.
From F.B.D. at A :
The correct answer is: 1440N
The spring block system lies on a smooth horizontal surface. The free end of the spring is being pulled towards right with constant speed v0 = 2m/s. At t = 0 sec, the spring of constant k = 100 N/cm in unstretched and the block has a speed 1m/s to left. The maximum extension of the spring is :
In the frame (inertial w.r.t. earth) of free end of spring, the initial velocity of block is 3 m/s to left and the spring unstretched.
Applying conservation of energy between initial and maximum extension state.
The correct answer is: 6cm
Block A is hanging from a vertical spring and is at rest. Block B strikes the block A with velocity v and sticks to it. Then the value of v for which the spring just attains natural length is :
The initial extension in spring is
Just after collision of B with A the speed of combined mass is v/2.
For the spring to just attain natural length the combined mass must rise up by (see fig.) and comes to rest.
Applying conservation of energy between initial and final states.
Alternate solution by SHM