Work, Energy and Power
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A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2 rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2 rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as
a = k2rt2
? tangential force, Ft = mat = mkr
Note ® Power delivered by centripetal force will be zero.
In the given figure, the block of mass m is dropped from the point ‘A’. The expression for kinetic energy of block when it reaches point ‘B’ is
In the given figure, the block of mass m is dropped from the point ‘A’. The expression for kinetic energy of block when it reaches point ‘B’ is
According to conservation of energy, we can write
Gain in kinetic energy = Loss in potential energy
Kf – Kin = Uin - Uf
K - = mgy – mg (y – y0) = mgy0
A 1kg particle is moving unidirectionally on a horizontal plane under the action of power , of power supplying energy source. The velocity (v) - time (t) graph that describes the motion of the particle is shown in figure. (Graphs are drawn schematically and are not to scale) The value of
in graph is
sec. Find 8x
A 1kg particle is moving unidirectionally on a horizontal plane under the action of power , of power supplying energy source. The velocity (v) - time (t) graph that describes the motion of the particle is shown in figure. (Graphs are drawn schematically and are not to scale) The value of
in graph is
sec. Find 8x
A particle of mass 2 kg is moving in the direction of positive x-axis. Initial velocity of the particle is 1 m/s. The instantaneous power vs time graph of the particle is shown in the figure. Find the velocity of the particle at 7 seconds.
A particle of mass 2 kg is moving in the direction of positive x-axis. Initial velocity of the particle is 1 m/s. The instantaneous power vs time graph of the particle is shown in the figure. Find the velocity of the particle at 7 seconds.
ΔW = area under P — t graph
= (1/2) (4 + 6) × 7 = 35 J
Work done = change in KE ⇒ 35 = (1/2) × 2 × v² - (1/2) × 2 × (1)² ⇒ v = 6 m/s
In an imaginary electric field, the electric potential energy of an electron at a distance r from the centre of force field is given by U = kr2, where k is a positive constant of appropriate dimensions. If the electron is moving in a circular orbit of radius r about the centre, then the orbital time period of the electron is proportional to:
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In an imaginary electric field, the electric potential energy of an electron at a distance r from the centre of force field is given by U = kr2, where k is a positive constant of appropriate dimensions. If the electron is moving in a circular orbit of radius r about the centre, then the orbital time period of the electron is proportional to:
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U = kr² ⇒ F = -dU/dr = -2kr ; 2kr = mv²/r ⇒ v = √ (2k/m) or T = 2πr/v = 2π√ (m/2k)
A smooth spherical ball strikes a smooth horizontal floor at an angle . The coefficient of restitution between the ball and the floor is . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)
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A smooth spherical ball strikes a smooth horizontal floor at an angle . The coefficient of restitution between the ball and the floor is . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)
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Vertical component of velocity just after collision
Fraction
What minimum speed does a 100g particle need At point B to reach point A? The graph shows potential energy versus position.
What minimum speed does a 100g particle need At point B to reach point A? The graph shows potential energy versus position.
To reach at point A particle must cross the peak point.
Loss in KE = gain in PE
v2 = 100
v = 10 m/s
Figure shown two blocks A and B having mass 2kg and 4kg moving with a speed 4 m/sec and 2 m/sec respectively. The maximum compression in the spring and final velocity of 2kg block and respectively.
Figure shown two blocks A and B having mass 2kg and 4kg moving with a speed 4 m/sec and 2 m/sec respectively. The maximum compression in the spring and final velocity of 2kg block and respectively.
From conservation of momentum
2 × 4 + 4 × 2 = 2v1 + 4v2
8 = v1 + 2v2
A plank of mass 5 kg in kept on a smooth surface. A block of 1 kg is kept on it and is given a velcoity of 12 m/s. There is friction between the block and the plank. Assuming the plank to be very long, find the magnitude of work done (in Joules) by friction on block till the relative sliding between them stops.
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A plank of mass 5 kg in kept on a smooth surface. A block of 1 kg is kept on it and is given a velcoity of 12 m/s. There is friction between the block and the plank. Assuming the plank to be very long, find the magnitude of work done (in Joules) by friction on block till the relative sliding between them stops.
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When relative sliding stops, both move with same velocity.
Only friction acts in horizontal direction.
∴ for system no external force, we can apply conservation of momentum.
1 × 12 + 5 × 0 = 6 × v
v = 2 m/s
Now work done by friction on block = change in kinetic energy of block
w = (1/2)m (2² - 12²) = (1/2
An inclined plane in bent in such a way that the vertical cross-section is given by y <!-- [if gte mso 9]><xml>
</xml><![endif]--> where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction <!-- [if gte mso 9]><xml>
</xml><![endif]-->= 0.5, the maximum height in cm at with a stationary block with not skip downward is ____________cm.
An inclined plane in bent in such a way that the vertical cross-section is given by y <!-- [if gte mso 9]><xml>
If the block does not slide,
mg sin
Thus,









