Ref: https://edurev.in/question/704041/Needed-a-Document-for-Work-power-energy-Related-Numerical-Questions-WORK-ENERGY
Here θ is the angle between and .
Work done by the force is positive if the angle between force and displacement is acute (0°<θ<90°) as cos θ is positive. This signifies, when the force and displacement are in same direction, work done is positive. This work is said to be done upon the body.
If applied force F is not a constant force, then work done by this force in moving the body from position A to B will be,
Here ds is the small displacement.
1J = 1 N.m , 1 erg = 1 dyn.cm
P = W/t = F.s/v = F.v
Here s is the distance and v is the speed.
1) Energy is the ability of the body to do some work. The unit of energy is same as that of work.
2) Kinetic Energy (K):- It is defined as,
K= ½ mv2
Here m is the mass of the body and v is the speed of the body.
Here, m is the mass of the body, g is the free fall acceleration (acceleration due to gravity) and h is the height.
In symbol’s,
U = mgh
K = p2/2m
W = ½ mv2 – ½ mu2
= Final K.E – Initial K.E
?
So, mgh + ½ mv2 = constant
In an isolated system, the total energy Etotal of the system is constant.
So, E = U+K = constant
Or, Ui+Ki = Uf+Kf
Or, ?U = -?K
Speed of particle v in a central force field:
v = √2/m [E-U(x)]
?In an isolated system (no external force ( Fext = 0)), the total momentum of the system before collision would be equal to total momentum of the system after collision.
So, pf = pi
e = relative velocity after collision / relative velocity before collision
= v2 – v1/u1 – u2
Case (i) For perfectly elastic collision, e = 1. Thus, v2 – v1 = u1 – u2. This signifies the relative velocities of two bodies before and after collision are same.
Case (ii) For inelastic collision, e<1. Thus, v2 – v1 < u1 – u2. This signifies, the value of e shall depend upon the extent of loss of kinetic energy during collision.
Case (iii) For perfectly inelastic collision, e = 0. Thus, v2 – v1 =0, or v2 = v1. This signifies the two bodies shall move together with same velocity. Therefore, there shall be no separation between them.
?
After collision, the velocity of two body will be,
v1 = (m1-m2/ m1+m2)u1 + (2m2/ m1+m2)u2
and
v2 = (m2-m1/ m1+m2)u2 + (2m1/ m1+m2)u1
Case:I
When both the colliding bodies are of the same mass, i.e., m1 = m2, then,
v1 = u2 and v2 = u1
Case:II
When the body B of mass m2 is initially at rest, i.e., u2 = 0, then,
v1 = (m1-m2/ m1+m2)u1 and v2 = (2m1/ m1+m2)u1
(a) When m2<<m1, then, v1 = u1 and v2 = 2u1
(b) When m2=m1, then, v1 =0 and v2 = u1
(c) When m2>>m1, then, v1 = -u1 and v2 will be very small.
v = (m1u1+m2u2) /(m1+m2)
and
loss in kinetic energy, E = ½ m1u12+ ½ m2u22 - ½ (m1+ m2)v2
or,
E= ½ (m1u12 + m2u22) – ½ [(m1u1+ m2u2)/( m1+ m2)]2
= m1 m2 (u1-u2)2 / 2( m1 + m2)
(i) The maximum transfer energy occurs if m1= m2
(ii) If Ki is the initial kinetic energy and Kf is the final kinetic energy of mass m1, the fractional decrease in kinetic energy is given by,
Ki – Kf / Ki = 1- v12/u21
Further, if m2 = nm1 and u2 = 0, then,
Ki – Kf / Ki = 4n/(1+n)2
(i) Momentum – m1u1+m2u2 = m1v1+m2v2
(ii) Energy – ½ m1u12+ ½ m2u22 = ½ m1v12+ ½ m2v22
So, F = - (dV/dr)
So,
Es = ½ kx2
Here k is the spring constant and x is the elongation.
(a) Condition for equilibrium, dU/dx = 0
(b) For stable equilibrium,
U(x) = minimum,
dU/dx = 0,
d2U/dx2 = +ve
(c) For unstable equilibrium,
U(x) = maximum
dU/dx = 0
d2U/dx2 = -ve
(d) For neutral equilibrium,
U(x) = constant
dU/dx = 0
d2U/dx2 = 0
UNITS AND DIMENSIONS OF WORK, POWER AND ENERGY
Work and Energy are measured in the same units. Power, being the rate at which work is done, is measured in a different unit.
Conversions between Different Systems of Units
1 Joule = 1 Newton ´ 1 m = 105 dyne ´ 102 cm = 107 erg
1 watt = 1 Joule/ sec = 107 erg/sec.
1 kwh = 103 watt ´ 1 hr = 103 watt ´ 3600 sec
= 3.6 ´ 106 Joule
1HP = 746 watt.
1 MW = 106 watt.
1 cal = 1 calorie = 4.2 Joule
1eV = "e" Joule = 1.6 ´ 10-19 Joule
(e = magnitude of charge on the electron in coulombs)
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