Work Power & Energy

$\mathrm{KE}_{\text{final }}=\dfrac{1}{2} m v^{2} \cos ^{2} 30^{\circ}=\dfrac{1}{2} m v^{2}\left(\dfrac{\sqrt{3}}{2}\right)^{2}$ $\dfrac{\mathrm{KE}_{\mathrm{in}}}{\mathrm{KE}_{\mathrm{f}}}=\dfrac{\dfrac{1}{2} m v^{2}}{\dfrac{1}{2} m v^{2}\left(\dfrac{3}{4}\right)}=\dfrac{4}{3}$

Given Restoring force, F = ax + bx2 Work done in stretching the rubber-band by a distance

is

Intergrating both sides,

=

=

When we extend the spring by

then the work done

Applying integration both sides we get, $\therefore$

The momentum $p$ of a particle is given by the product of its mass $m$ and its velocity $v$, that is, $p = m \cdot v$.

For a given momentum, the relationship between mass and velocity can be understood as inversely proportional.

This means that as the mass increases, the velocity decreases to maintain the same momentum, and vice versa.

The kinetic energy ($K.E.$) of a particle is given by the formula $K.E. = \dfrac{1}{2} m v^2$.

This equation shows that the kinetic energy depends on both the mass of the particle and the square of its velocity.

Given that four particles $A, B, C, D$ have masses $\dfrac{m}{2}, m, 2 m, 4 m$, respectively, and all have the same momentum, we can assume the momentum of each particle to be $p$.

This common value of momentum allows us to express the velocity of each particle in terms of its mass and the common momentum $p$.

The velocity $v$ of each particle will be $v = \dfrac{p}{m}$.

Thus, for each particle, we can determine the velocity as follows: For $A$: $v_A = \dfrac{p}{\dfrac{m}{2}} = \dfrac{2p}{m}$ For $B$: $v_B = \dfrac{p}{m}$ For $C$: $v_C = \dfrac{p}{2m} = \dfrac{p}{2m}$ For $D$: $v_D = \dfrac{p}{4m}$ Now, substituting these velocities into the kinetic energy formula yields the kinetic energies for each particle: $K.E._A = \dfrac{1}{2} \cdot \dfrac{m}{2} \cdot \left(\dfrac{2p}{m}\right)^2 = \dfrac{1}{2} \cdot \dfrac{m}{2} \cdot \dfrac{4p^2}{m^2} = \dfrac{2p^2}{m}$ $K.E._B = \dfrac{1}{2} \cdot m \cdot \left(\dfrac{p}{m}\right)^2 = \dfrac{1}{2} \cdot m \cdot \dfrac{p^2}{m^2} = \dfrac{p^2}{2m}$ $K.E._C = \dfrac{1}{2} \cdot 2m \cdot \left(\dfrac{p}{2m}\right)^2 = \dfrac{1}{2} \cdot 2m \cdot \dfrac{p^2}{4m^2} = \dfrac{p^2}{4m}$ $K.E._D = \dfrac{1}{2} \cdot 4m \cdot \left(\dfrac{p}{4m}\right)^2 = \dfrac{1}{2} \cdot 4m \cdot \dfrac{p^2}{16m^2} = \dfrac{p^2}{8m}$ Comparing these kinetic energies, we see that the particle $A$ has the maximum kinetic energy, as it is inversely related to mass in this scenario, and $A$ has the least mass but the highest velocity squared component, thus maximizing its kinetic energy.

Therefore, the correct answer is: Option D: A

Let

be the velocity with which the particle is thrown and

be the mass of the particle. Then

At the highest point the velocity is

(only the horizontal component remains, the vertical component being zero at the top-most point).

Therefore kinetic energy at the highest point,

[ From eq

]

Work done =

$=$ 12 J work energy theorem wnet $=$

K.E. 12 $=$ Kf $-$ 3 Kf = 15 J

When a body moves under the influence of a constant power, the relationship between displacement and time can be established through the concept of power.

Power (P) is defined as the rate at which work is done, and it can also be expressed in terms of force (F) and velocity (v) as $P = F \cdot v$.

For a constant power P and assuming the force acts in the direction of the velocity, we can analyze how displacement (s) changes with time (t).

Since force can also be written as $F = \dfrac{d(mv)}{dt}$ for a constant mass m, this simplifies to $F = m \dfrac{dv}{dt}$, because mass doesn't change with time for most cases.

Integrating force over a distance gives work (W), and power is the rate of doing work, thus we can connect these concepts.

The kinetic energy (K.E) of the body is given by $K.E = \dfrac{1}{2}mv^2$, and the work done by the force is equal to the change in kinetic energy.

Considering power is constant, $P = \dfrac{dW}{dt} = \dfrac{d(\dfrac{1}{2}mv^2)}{dt}$.

Rearranging terms to focus on velocity and integrating with respect to time will give us a relation involving velocity and time.

For a constant mass system, and using $P = F \cdot v = m \cdot a \cdot v = m \cdot \dfrac{dv}{dt} \cdot v$, and knowing that $P = \text{constant}$, we rearrange to find the relationship between velocity and time.

Given $P = m \cdot v \cdot \dfrac{dv}{dt}$, we rearrange to $\dfrac{P}{m} dt = v dv$.

Integrating both sides where the initial condition is when $t = 0, v = 0$, we get $\dfrac{P}{m} t = \dfrac{1}{2} v^2$, solving for $v$ gives $v \propto t^{1/2}$, so $v = k \cdot t^{1/2}$ for some constant $k$.

The displacement $s$ is obtained by integrating the velocity with respect to time, $s = \int v dt = \int k \cdot t^{1/2} dt = \dfrac{2}{3}k \cdot t^{3/2}$.

Therefore, the displacement $s$ is proportional to $t^{3/2}$.

The correct answer is Option B, $t^{3/2}$.

We know that

Power According to the question, power is constant. $\therefore$

where

constant $\therefore$

$\therefore$

$\therefore$

$\therefore$

$\therefore$

where

$\therefore$

Assume the amount of fat is used = x kg So total Mechanical energy available through fat =

And work done through lifting up = 10 $\times$ 9.8 $\times$ 1000 = 98000 J $\Rightarrow$

= 98000 $\Rightarrow$

= 12.89 $\times$ 10-3 kg

The correct answer is:

Here's why: Conservative vs.

Non-Conservative Forces: Conservative Forces: These are forces for which the work done is independent of the path taken; they only depend on the initial and final positions.

For conservative forces, you can define a potential energy function.

Examples include Coulomb’s force, gravitational force, and the restoring force of a spring.

Non-Conservative Forces: These forces depend on the path taken and usually result in energy dissipation (often as heat).

Friction is a typical example, and hence, it cannot be described solely by a potential energy function.

Example Explanations: Coulomb’s Force: It is a conservative force because the work done in moving a charge in an electric field only depends on the difference in electric potential between two points.

Gravitational Force: It is also conservative.

The potential energy associated with gravitational force is given by gravitational potential energy, for example,

near the Earth’s surface. Restoring Force: For a spring, the force is given by Hooke’s law:

and the corresponding potential energy is

This is a conservative force.

Since friction is a non-conservative force that dissipates energy, it cannot be expressed in terms of potential energy.

Therefore, the correct answer is Option A: Frictional force.