Work, Energy and Power

6.5 (a) As per the law of conservation of energy,

Total energy = potential energy + kinetic energy

= mgh + $\frac{1}{2}$ mv2

When the casing burns, mass reduces, resulting in drop of energy. Hence the energy for burning of casing is drawn from the rocket.

(b) The force due to gravity is a conservative force. The work done on a closed path for a conservative force is zero. Hence, for every complete orbit of the comet, the work done by the gravitational force is zero.

(c) The potential energy of the satellite revolving the Earth decreases as it approaches the Earth and since the system's total energy should remain constant, the kinetic energy increases. Hence the satellite velocity increases when it approaches the Earth.

(d)

(i) Give, m = 15 kg, displacement, s = 4m

From the relation, work done, W = F $*s*\cos ??$ where $?$ is the angle between the force and displacement. Here F = mg, hence, W = mgs $\cos ??$ = 15 $*9.8*4*\cos ?90°$ = 0

(ii) Mass = 15 kg, s = 4 m, the applied force direction is same as the direction of the displacement, $?=0$ . Hence W = 15 $*9.8*4*\cos ?0°$ = 588 J

6.3 We know the total energy E is given by E = Kinetic energy (KE) + Potential energy (PE)

(a) In figure (a), we have at x=0, the potential energy is zero. So KE is positive. At x>a, the potential energy has a value greater than E, so the KE becomes 0. Thus the particle will not exist in the region x>a. Minimum total energy is zero.

(b) For the entire x-axis, PE >E, the KE of the object would be negative. Thus the particle will not exist in this region.

(c) In x=0 to x=a and x>b, PE is greater than E, so he KE has to be negative. The object cannot exist in this region.

(d) For x=-b/2 to x =-a/2 and x=a/2 to x=b/2, KE is positive and the PE

6.2 Given, mass of the body, m = 2 kg

Horizontal force applied, F = 7 N

Coefficient of friction, $?$ = 0.1

Acceleration, a = F/m = 7/2 = 3.5 m/s2

Frictional force, f = $?R=?mg=$ 0.1 $*2*9.807=$ 1.96 N

Retardation produced by the frictional force, $a_r$ = -f/m = -1.96 /2 = 0.98 m/s2

The net acceleration by which the body moves forward

$a_n$ = a - $a_r$ = 3.5 – 0.98 = 2.52 m/s2

Distance moved by the body in 10 s is given by

s = ut + (1/2) $a_n{t}^2$ = 0 $*10+\left(\frac{1}{2}\right)*2.52*{10}^2$ = 126 m

(a) Work done in 10 s is given by

W = Force $*\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}=7*126=882\mathrm{J}$

(b) Work done by friction in 10 s is given by

W = -f $*s=-1.96*126=$ - 247J

(c) Work done by the net force,

W = (F-f) $*s$ = (7 – 1.96 ) $*126$ = 635 J

(d) Kinetic energy is given by the equation

KE = (1/2)m $v^2$ where v is the final velocity

From the equation v = u + at we get after 10 s, v = 0 + 2.52 $*$ 10 = 25.2 m/s

Final kinetic energy = (1/2) $*2*{25.2}^2$ = 635 J

Initial kinetic energy = (1/2) $*2*0^2$ = 0

Change in kinetic energy = 635 – 0 = 635 J

Hence the work done by the net force = change in kinetic energy

6.1  (a) While the person lifts a bucket out of a well by means of a rope tied to the bucket, the direction of both the force and the displacement are same, hence the work done is positive.

(b) While lifting the bucket, he works against gravity, but the work done by the gravitational force is downward, hence the work done is negative.

(c) The direction of motion of the object is in the opposite direction of the frictional force; hence the work done is negative.

(d) While a body moves on a rough horizontal plane, the frictional forces try to oppose the motion. But since the applied force maintains uniform velocity of the object, the motion of the object and the applied force are in the same direction. Hence the work done is positive.

(e) Since the resistive force of air is trying to bring the vibrating pendulum to rest, the work done is negative.