The negative of the distance rate of change of potential energy is equal to

force acting on the particle in the direction of displacement
acceleration of the particle, perpendicular to d is placement
power
impulse

Solution

F = - ^dU⁄_dx

The force F acting on a body moving in a circle of radius r is always perpendicular to the instantaneous velocity v.The work done by the force on the body in half rotation is

Fv
F·2πr
Fr

Solution

Work done by centripetal force is zero.

A ramp is constructed in parabolic shape such that the height y of any point on its surface is given in terms of the point’s horizontal distance x from the bottom of the ramp bey = x²/2L.A block of granite is to be set on the ramp; the coefficient of static friction is 0.80. What is the maximum x coordinate at which the block can be placed on the ramp and remain at rest, if L = 10 m?

8 m
8.4 m
9 m
9.4 m

Solution

The kinetic energy of particle moving along a circle of radius R depends upon the distance and is given by K= aS where a is a constant. Then the force acting on the particle is

\(\frac{aS}{R}\)
\(\frac{2(aS)^{2}}{R}\)
\(\frac{aS^{2}}{R^{2}}\)
\(\frac{2aS}{R}\)

Solution

A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest.After collision, their final velocities are V and v respectively.The value of v is

\(\frac{2uM}{m}\)
\(\frac{2um}{M}\)
\(\frac{2u}{1+\frac{m}{M}}\)
\(\frac{2u}{1+\frac{M}{m}}\)

Solution

A 10 m long iron chain of linear mass density 0.8 kg m^-1 is hanging freely from a rigid support. If g = 10 ms^-2, then the power required to left the chain upto the point of support in 10 second

10 W
20 W
30 W
40 W

Solution

A weight suspended from the free end of a vertically hanging spring produces an extension of 3 cm. The spring is cut into two parts so that the length of the longer part is ²⁄₃ of the original length, If the same weight is now suspended from the longer part of the spring, the extension produced will be

0.1 cm
0.5 cm
1 cm
2 cm

Solution

A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest is

eⁿt
e²t
\(t\left [ \frac{1+e}{1-e} \right ]\)
\(t\left [ \frac{1-e}{1+e} \right ]\)

Solution

A motor drives a body along a straight line with a constant force. The power P developed by the motor must vary with time t according to

Solution

P = F × v ⇒ P =F at ∴ P ∝ t

A mass m is moving with velocity v collides inelastically with a bob of simple pendulum of mass m and gets embedded into it. The total height to which the masses will rise after collision is

^V²⁄_8g
^V²⁄_4g
^V²⁄_2g
^2V²⁄_g

Solution

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