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The minimum velocity (in ms^{1}) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is
[AIEEE 2002]
To move a body on a circular path, centripetal force is necessary. In this question, there is an agent, friction that will provide the required centripetal force to the car.
Using the relation,
mv^{2}r=μR,R=mg
mv^{2}r=μmg
or, v^{2}=μrg
or, v^{2}=0.6×150×10
or, v=30m/s
Which of the following statements is false for a particle moving in a circle with a constant angular speed ?
[AIEEE 2004]
In uniform circular motion tangential acceleration is zero only centripetal acceleration work which is directed towards centre.
An angular ring with inner and outer radii R_{1} and R_{2} is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, F_{1}/F_{2} is
F = mω^{2}R
Since F α R, the ratio of forces will be equal to the ratio of radii.
Hence D is the correct answer.
A point P moves in counterclockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length s = t^{3} + 5, where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t = 2s is nearly
[AIEEE 2010]
= 14m/s^{2}
For a particle in uniform circular motion the acceleration at a point P(R,) on the circle of radius R is (here is measured from the xaxis)
[AIEEE 2010]
In a typical power station, amid a series of energy transfers and conversions, about 70% of energy input gets wasted in form of thermal energy, efficiency of such a power station is about
Assume total input energy to be 100%
Lost energy =70%
Left energy = 100 – 70 = 30 % is the output energy
Efficiency = output energy/input energy = 30/100 = 0.3
Percentage efficiency = 0.3 × 100 = 30%
A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
[AIEEE 2002]
The work done is stored as elastic potential energy in the spring. It is given by
W = 1/2 * k (x₂^{2}  x₁^{2})
x₁ = 5 cm
x₂ = 15 cm
Force constant, k = 800 N/m = 800/100 N/cm = 8 N/cm
W = 1/2 * 8 (15^2  5^2) = 1/2 * 8 (225  25)
W = 1/2 * 8 * 200 = 800 N cm
W = 800 N cm = 8 N m = 8 J
Therefore the work done in extending it from 5cm to 15cm is 8 J.
Hence B is the correct answer.
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to
[AIEEE 2003]
P= fv
P = mav
p = mdv/dt.v
p.dt/m = v.dv ( integration from 0t)
pt/m = v²/2
v = √(2pt/m)
ds/dt = t½dt
s = t^{3/2}
A spring of spring constant 5 × 10^{3} N/m is stretched initally by 5 cm from the unstreched position. Then the work required to stretch it further by another 5 cm is
[AIEEE 2003]
work done in stretching a spring by a distance x is
W=½kx^{2}
where k is the spring constant.
work required to stretch the string by 10cm is
W_{1}=½kx^{2}=½ 5×103(0.1)^{2}=25J
work required to stretch the string by 5cm is
W_{2}=½kx^{2}=½ 5×10^{3}(0.05)^{2}=6.25J
Hence the work done in stretching it from 5cm to 10cm is
W_{1}−W_{2}=25−6.25=18.75J
A force F = (5i + 3j +2 k) N is applied over a particle which displaces it from its origin to the point r = (2 i  j) m. The work done on the particle in joule is
[AIEEE 2004]
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
[AIEEE 2004]
The center of mass of the hanging part is at 0.3m from table
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
[AIEEE 2004]
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is
[AIEEE 2005]
A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rools down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball i
[AIEEE 2005]
mgh = 1/2×mv^{2}
v = √2gh
v = √2×10×80(actually 80 is come out of from 10020)
v = 40 m/s
Hence A is correct.
A body of mass m is accelerated uniformly from rest to a speed V in a time T. The instantaneous power delivered to the body as function of time, is given by
[AIEEE 2005]
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion ?
[AIEEE 2005]
A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is
[AIEEE 2006]
By the Work Energy Theorem
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s^{2 }
[AIEEE 2006]
The potential energy of a 1 kg particle free to move along the xaxis is given by
The total mechanical energy of the particle is 2 J. Then, the maximum speed (in ms^{_1}) is
[AIEEE 2006]
A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10000 N/m. The spring compresses by
[AIEEE 2007]
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
[AIEEE 2008]
At time t = 0 s particle starts moving along the xaxis. If its kinetic energy increase uniformly with time t, the net force acting on it must be proportional to
[AIEEE 2011]
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