Work, Power & Energy

Since the car is moving with a constant velocity, so
Hence, the correct answer is (A).

When it is lowered gradually then to attain the equilibrium, we have
$m g = k d$...(i)
Further when it is allowed to fall suddenly, and if x is the maximum stretching in the spring, then by Law of Conservation of Energy
$\begin{array}{l} \left( \begin{array}{c} \text {Loss in Gravitational} \\ \text {Potential Energy} \\ \text {of Body} \end{array} \right) = \left( \begin{array}{c} \text {Gain in Elastic} \\ \text {Potential Energy} \\ \text {of Spring} \end{array} \right) \\ \Rightarrow m g x = \frac {1}{2} k x ^ {2} \\ \Rightarrow m g = \frac {1}{2} k x \end{array}.....(ii)$
Equating (1) and (2), we get
x = 2 d
Hence, the correct answer is (B).

By Law of Conservation of Energy
$\begin{array}{l} \frac {1}{2} m v ^ {2} = \frac {1}{2} k x ^ {2} \\ \Rightarrow \quad \frac {1}{2} (4) (1.6) ^ {2} = \frac {1}{2} (2) x ^ {2} \\ \Rightarrow \quad x = 2.2 6 \mathrm{m} \\ \text { Hence,the correct answer is(C). } \end{array}$

$a = g (\sin 3 0 - \mu \cos \theta)$
$ \Rightarrow a = 9.8 \left(\frac {1}{2} - 0.2 \frac {\sqrt {3}}{2}\right)$
$\Rightarrow a = 9.8 (0.3 2 6 8)$
$\Rightarrow a = 3. 2 \mathrm{ms} ^ {- 2}$
$ \text {Since} v ^ {2} - (1 0) ^ {2} = 2 (3.2) (1 0) $
$\Rightarrow v ^ {2} = 1 6 4 $
$\Rightarrow v \cong 1 3 \mathrm{ms} ^ {- 1} $
Hence, the correct answer is (B).

Speed will be maximum at equilibrium, i.e., when
$\begin{array}{r l} & F = k x \\ \Rightarrow & x = \frac {F}{k} \\ \text {Applying MWET, we get} \\ & W _ {e x t} = \Delta U + \Delta K \\ \Rightarrow & F x = \frac {1}{2} k x ^ {2} + \frac {1}{2} m v ^ {2} \\ \Rightarrow & F \left(\frac {F}{k}\right) = \frac {1}{2} k \left(\frac {F}{k}\right) ^ {2} + \frac {1}{2} m v ^ {2} \\ \Rightarrow & \frac {1}{2} m v ^ {2} = \frac {F ^ {2}}{2 k} \\ \Rightarrow & v = \frac {F}{\sqrt {m k}} \end{array}$
Hence, the correct answer is (A).

$\begin{array}{r l} & {\binom {\mathrm{LossinKinetic}} {\mathrm{Energy}} = \binom {\mathrm{GaininGravitational}} {\mathrm{PotentialEnergy}}} \\ {\Rightarrow} & {\frac {1}{2} m V _ {0} ^ {2} - \frac {1}{2} m V ^ {2} = m g h} \\ {\Rightarrow} & {V ^ {2} = V _ {0} ^ {2} - 2 g h} \\ {\Rightarrow} & {V = \sqrt {V _ {0} ^ {2} - 2 g h}} \end{array}$
Hence, the correct answer is (C).

Potential energy of cube at position 1,
$U _ {1} = M g (4 R) = 4 M g R$
Applying Law of Conservation of Energy between 1 and 2, we get
$\begin{array}{r l} & K _ {1} + U _ {1} = K _ {2} + U _ {2} \\ \Rightarrow & 4 M g R = 2 M g R + \frac {1}{2} M v ^ {2} \\ \Rightarrow & 4 M g R = M v ^ {2} \\ \Rightarrow & v ^ {2} = 4 g R \\ \mathrm{Since,} N = \frac {M v ^ {2}}{R} - M g \\ \Rightarrow & N = \frac {M}{R} (4 R g) - M g = 3 M g \end{array}$
Hence, the correct answer is (A).

Let $l$ be the length of spring at the situation shown in Figure.
$\begin{array}{r l} \Rightarrow & \cos (3 7 ^ {\circ}) = \frac {h}{l} \\ \Rightarrow & l = \frac {h}{\cos (3 7 ^ {\circ})} = \frac {5 h}{4} \end{array}$
Extension $x$ in spring is
$x = l - h = \frac {5 h}{4} - h = \frac {h}{4}$
By Law of Conservation of Energy, we have
$\begin{array}{r l} & {\binom{\text {Loss in EPE}}{\text {of springs}} = \binom{\text {Gain in KE}}{\text {of Bead}}} \\ {\Rightarrow} & {\frac {1}{2} k x ^ {2} + \frac {1}{2} k x ^ {2} = \frac {1}{2} m v ^ {2}} \\ {\Rightarrow} & {2 \bigg (\frac {1}{2} (1 0 0 0) \frac {h ^ {2}}{1 6} \bigg) = \frac {1}{2} (5) v ^ {2}} \\ {\Rightarrow} & {v ^ {2} = 2 5 h ^ {2}} \\ {\Rightarrow} & {v = 5 h} \end{array}$
Hence, the correct answer is (A).

Speed of ball is maximum at equilibrium, so
$x _ {e q} = \frac {m g}{k}$
Hence, the correct answer is (A).

According to Work-Energy Theorem for a non-conservative system, we have
$W _ {c} + W _ {n c} = \Delta k$
{OPTION (C)}
Also, we know that the work done by a conservative force equals the decrease in potential energy, so
$\begin{array}{r l} & W _ {c} = - \Delta U \\ \Rightarrow & W _ {n c} - \Delta U = \Delta k \end{array}\tag{{OPTION (B)}}$
{OPTION (A)}
Hence, (A), (B) and (C) are correct.

Tension in the string will be maximum when mass m is at point B, so
$T = m g + \frac {m v ^ {2}}{\ell}$
By Law of Conservation of Energy,
$\begin{array}{r l} & {\frac {1}{2} m v ^ {2} = m g h = m g \ell (1 - \cos \theta)} \\ {\Rightarrow} & {v ^ {2} = 2 g \ell (1 - \cos (6 0 ^ {\circ}))} \\ {\Rightarrow} & {T = m g + \frac {2 m g \ell}{\ell} \bigg (1 - \frac {1}{2} \bigg)} \\ {\Rightarrow} & {T = 2 m g} \\ {\mathrm{Formass4mtobestationary,} f \geq T} \\ {\Rightarrow} & {\mu (4 m g) \geq 2 m g} \\ {\Rightarrow} & {\mu \geq 0. 5} \end{array}$
Hence, (A) is correct
When m moves from A to B it has vertical displacement downwards, i.e., along the direction of gravitational force.
So, B) is correct.
Tension is always perpendicular to the velocity, so power delivered by the tension is zero.
So, C) is correct.
According to Work Energy Theorem,
$W _ {\mathrm{total}} = \Delta K$
Since work done by tension is zero therefore work is done only by the gravitational force. So, (D) is also correct.
Hence, (A), (B), (C) and (D) are correct.

$P = (\text {Total Frictional Force}) (\text {velocity})$
$\Rightarrow P = (m f) v $
$\Rightarrow P _ {\text {extra}} = m g (\sin \theta) v = \frac {m g h v}{s} $
Hence, (A), (B) and (C) are correct.

Using the concepts of Laws of Motion and Kinematics we can draw following diagrams and then using the definition of work, we can find corresponding work done.

Hence, (A), (B), (C) and (D) are correct.

For a light string pulled by a constant
Force $F = 40 \mathrm{~N}$ , tension (T) in the string is also $40 \mathrm{~N}$ . So, $T = 40 \mathrm{~N}$
Since $W = 40 \mathrm{~J} = Fs$ where $F = 40 - 20 \mathrm{~N} = 20 \mathrm{~N}$
$\begin{array}{r l} \Rightarrow & 4 0 = 2 0 \mathrm{s} \\ \Rightarrow & s = 2 \mathrm{m} \end{array}$
Since acceleration is constant, so $v \propto t$ and hence $P = Fv$
$\Rightarrow P \propto t$
$\mathrm{Also}, v = \sqrt {2 a x}$
$\Rightarrow P \propto \sqrt {x}.$
So, all options (A), (B), (C) and (D) are correct.
Hence, (A), (B), (C) and (D) are correct.

Work done by $\vec{F}_1$ is
$W _ {1} = \int_ {P _ {1}} ^ {P _ {2}} F _ {1} \cos \theta d s$
Where, $ds = (6)d(-2\theta) = -12d\theta$ and $F_{1} = 20 N$
$\Rightarrow W _ {1} = - 2 4 0 \int_ {\pi / 4} ^ {0} \cos \theta d \theta$
$\Rightarrow W _ {1} = 2 4 0 \sin \left(\frac {\pi}{4}\right) = 1 2 0 \sqrt {2} \mathrm{J}$

$\vec{F}_{1}$ is conservative because it is always directed towards a fixed point $P_{2}$ . Therefore, $W_{1}$ can be directly calculated as
$W _ {1} = F _ {1} \left(P _ {1} P _ {2}\right) = (2 0) \left(6 \sqrt {2}\right) = 1 2 0 \sqrt {2} \mathrm{J}$
Similarly, $W_{2} = F_{2}(OP_{2}) = (30)(6) = 180 \mathrm{~J}$
and
$W _ {3} = \int_ {0} ^ {6 (\pi / 2)} F _ {3} d s = \int_ {0} ^ {3 \pi} 1 5 d s$
$W _ {3} = 1 5 s | _ {0} ^ {3 \pi} = 4 5 \pi \mathrm{J}$
Hence, (A), (C) and (D) are correct.

According to Work Energy Theorem, work done by all the forces is equal to change in kinetic energy, so
$\begin{array}{r l} & W _ {\mathrm{total}} = \Delta K \\ \Rightarrow & W _ {\mathrm{ext}} + W _ {n c} = \Delta U + \Delta K \end{array}$

Since, $W_{nc} = 0$ , because friction is absent
$\mathrm{So}, W _ {\mathrm{ext}} = \Delta U + \Delta K$
$\begin{array}{r l} \Rightarrow & (M a) x _ {\max} = \frac {1}{2} k x _ {\max} ^ {2} \\ \Rightarrow & x _ {\max} = \frac {2 M a}{k} \end{array}$
The block will execute SHM and its initial position will be the one of the extreme positions i.e., block returns to its initial position after maximum elongation in the spring and compression in the spring is zero.
So, (D) is also correct.
Hence, (B) and (D) are correct.

$\begin{array}{l} W = \int \vec {F} \cdot d \vec {r} = \int (\hat {i} F _ {x} + \hat {j} F _ {y}) \cdot (\hat {i} d x + \hat {j} d y) \\ W = \int (7 \hat {i} - 6 \hat {j}) \cdot (\hat {i} d x + \hat {j} d y) \\ \Rightarrow W = 7 \int_ {0} ^ {- 3} d x - 6 \int_ {0} ^ {4} d y = - 2 1 - 2 4 = - 4 5 J \end{array}$
Displacements along the direction of X and Y axes are opposite to the direction of force acting along the respective directions, hence the particle must have some velocity at $(0, 0)$ .
Hence, (B) and (D) are correct.

In region $AB$ , $s \propto t$
$\begin{array}{r l} \Rightarrow & v = \mathrm{constant} \\ \Rightarrow & \Delta K = 0 \\ \Rightarrow & W _ {A \to B} = \Delta K = 0 \end{array}$
Similarly, from B to C, slope of s-t graph is decreasing, so v is also decreasing and hence
$W = \Delta K = \mathrm{NEGATIVE}$
Hence, (B) and (C) are correct.

Let m be the mass of the block
Initial elongation of the spring will be
$x _ {i} = \frac {m g}{k}\tag{... (1}$
On application of F work done by F and gravity is used to increase the elastic potential energy of spring. So,
$ (F + m g \big) x _ {0} = \frac {1}{2} k (x _ {i} + x _ {0} ) ^ {2} - \frac {1}{2} k x _ {i} ^ {2}$....(2)
From 1) and 2) we get,
$x _ {0} = \frac {2 F}{k}$
Work done by applied force F is $W = Fx_{0}$
Hence, (C) and (D) are correct.

Since $P = \vec{F} \cdot \vec{v}$ . At 1 s and 3 s, P is positive. Hence, angle is acute. At 7 s power is negative. Hence, angle is obtuse.
Area under P-t graph is the work done which equals the change in kinetic energy.
Hence, (A), (C) and (D) are correct.

In a conservative field, the work done $W_{C}$ by the conservative force is always equal to the decrease in potential energy.
$\Rightarrow W _ {c} = U _ {\mathrm{initial}} - U _ {\mathrm{final}} = - \Delta U$
$\Rightarrow W _ {c} = U _ {A} - U _ {B}$
Decrease in potential energy implies $U_{A} > U_{B}$
Hence, (B) and (C) are correct.

Conceptual (B) and (D) are correct.

By Work-Energy Theorem
$\frac {1}{2} m v ^ {2} - \frac {1}{2} m u ^ {2} = \vec {F} \cdot \vec {s}$
K.E. increasing with time means v > u
$\Rightarrow$ Either $\theta = 0^{\circ}$
OR $0 < \theta \leq 90^{\circ}$ i.e. $\theta$ is acute
OR $P = \frac{W}{t}\neq 0$ , because
$W \neq 0$
Since, K.E. is increasing, so P.E. must be decreasing and hence height above the ground must be decreasing. Hence, (A), (C), (D) and (E) are correct.

Force on the particle is given by
$\begin{array}{r l} & {\vec {F} = - \nabla U} \\ {\Rightarrow} & {\vec {F} = - \frac {\partial U}{\partial x} \hat {i} - \frac {\partial U}{\partial y} \hat {j} = - (6 x y ^ {2} + 6) \hat {i}} \end{array}$
Now, for initial acceleration i.e. t = 0 and x = 1, y = 1, we have
$\begin{array}{r l} & {\vec {a} = \frac {\vec {F}}{M} = 1 2 \hat {i} - 6 \hat {j}} \\ {\Rightarrow} & {| \vec {a} | = 6 \sqrt {5} \mathrm{ms} ^ {- 2}} \end{array}$
Since particle is at rest at x=1, y=1 so, the total mechanical energy (E) is
$\begin{array}{r l} & E = U + K \\ \Rightarrow & U (1, 1) + K (1, 1) = E \\ \Rightarrow & E = 9 \mathrm{J} \end{array}$
According to modified work-Energy Theorem, we have
$W _ {\mathrm{ext}} = \Delta U + \Delta K$
Since the particle is moving slowly, so $\Delta K = 0$
$\begin{array}{r l} \Rightarrow & W _ {\mathrm{ext}} = \Delta U = U (0, 0) - U (1, 1) \\ \Rightarrow & W _ {\mathrm{ext}} = - 9 \mathrm{J} \end{array}$
Hence, (A) and (C) are correct.

Free body diagram of block is as shown in figure.

From Work Energy Theorem
$W _ {\mathrm{net}} = \Delta \mathrm{KE}$
$\Rightarrow (4 0 - 2 0) \mathrm{s} = 4 0$
$\Rightarrow s = 2 \mathrm{m}$
Work done by gravity is
$W _ {g} = - 2 0 \times 2 = - 4 0 \mathrm{J}$
and work done by tension is
$W _ {T} = 4 0 \times 2 = 8 0 \mathrm{J}$
Hence, (A), (B) and (D) are correct.