Work, Energy and Power

Assertion : True

Reason : False : As potential energy is associated to conservative forces only.

$$\begin{gathered} K_{·E}·=\frac{1}{2}m{u}^2 \\ \text{ At maximum height } \\ K·E_1=\frac{1}{2}m(u\cos \theta {)}^2 \\ K·{\epsilon }_1=\frac{1}{2}m{u}^2{\cos }^{2}60° \\ K·{\epsilon }_1=K·E·x\frac{1}{4} \\ K·{\epsilon }_1=\frac{K·E·}{4} \end{gathered}$$

Assertion : True

Reason : False : As potential energy is associated to conservative forces only.

$$\begin{gathered} V_{\text{Top }}=\sqrt{n^{2}gR} \\ V_{\text{Botom }}=\sqrt{n^{2}gR+4gR} \\ \text{ Ratio }=\frac{n^2+4}{n^2} \end{gathered}$$

Central force is conservative force

$$\begin{gathered} K_{·E}·=\frac{1}{2}m{u}^2 \\ \text{ At maximum height } \\ K·E_1=\frac{1}{2}m(u\cos \theta {)}^2 \\ K·{\epsilon }_1=\frac{1}{2}m{u}^2{\cos }^{2}60° \\ K·{\epsilon }_1=K·E·x\frac{1}{4} \\ K·{\epsilon }_1=\frac{K·E·}{4} \end{gathered}$$

Assertion : True

Reason : False : As potential energy is associated to conservative forces only.

Assertion : True

Reason : False : As potential energy is associated to conservative forces only.

1. Initial Kinetic Energy:

   • The total initial kinetic energy ($K$K) of the projectile is given as: $$K = \frac{1}{2} m v^2$$
   • where:
     • $m$ is the mass of the projectile.
     • $v$ is the initial velocity of the projectile.

2. Velocity Components:

   • The initial velocity ($v$v) can be resolved into horizontal and vertical components: $$v_x = v \cos \theta, \quad v_y = v \sin \theta$$
   • Here, $\theta = 60^\circ$

3. At the Highest Point:

   • At the highest point, the vertical velocity ($v_y$) becomes zero due to gravity, but the horizontal velocity ($v_x$) remains constant throughout the motion because there is no acceleration in the horizontal direction.

   • The velocity at the highest point is purely horizontal:

     $$v_\text{highest} = v_x = v \cos 60^\circ$$

     Since $\cos 60^\circ = \frac{1}{2}$

     $$v_\text{highest} = \frac{v}{2}$$​

4. Kinetic Energy at the Highest Point:

   • The kinetic energy at the highest point is due only to the horizontal velocity: $$K_\text{highest} = \frac{1}{2} m v_\text{highest}^2$$
   • Substituting $v_\text{highest} = \frac{v}{2}$
   • $$K_\text{highest} = \frac{1}{2} m \left(\frac{v}{2}\right)^2$$
   • $$K_\text{highest} = \frac{1}{2} m \cdot \frac{v^2}{4}$$
   • ​ $$K_\text{highest} = \frac{1}{4} \left(\frac{1}{2} m v^2\right)$$
   • Since $K = \frac{1}{2} m v^2$,
   • we have: $$K_\text{highest} = \frac{K}{4}$$​

The kinetic energy of the projectile at the highest point is:

Correct Option: C. $K/4$

The work done by the force is given by:

where $d\vec{r} = dx \hat{i} + dy \hat{j}$​. Along $x + y = 10$, we use $y = 10 - x$ and $dy = -dx$, resulting in:

Integrating from $x = 0$ to $x = 4$:

After solving, we find:

Thus, the work done by the force is $-112 \, \text{J}$

the maximum velocity for a simple harmonic oscillator is given by:

$V=A\omega$

The angular frequency $\omega$ of a mass-spring system is related to the spring constant k and the mass m by:

$\omega =\sqrt{\frac{k}{m}}$

Since the bodies have equal amplitudes, therefore,

$$\begin{gathered} \frac{V_1}{V_2}=\frac{{\omega }_1}{{\omega }_2} \\ \frac{V_1}{V_2}=\frac{\sqrt{\frac{k_1}{m}}}{\sqrt{\frac{k_2}{m}}} \\ \frac{V_1}{V_2}=\sqrt{\frac{k_1}{k_2}} \end{gathered}$$

The work done by this force when the object moves from x=0 to x=1 meter is given by the integral of the force over the displacement:

$$\begin{gathered} \text{ Work done }={\int }_0^1\left(\alpha +\beta x^2\right)dx \\ 5={\int }_0^1\left(\alpha +\beta x^2\right)dx \\ 5={\left[\alpha x+\frac{\beta x^3}{3}\right]}_0^1 \end{gathered}$$

After evaluating:

$5=\alpha +\frac{\beta }{3}$

Given that a=1 N, we substitute this value:

$$\begin{gathered} 5=1+\frac{\beta }{3} \\ \frac{\beta }{3}=4\Rightarrow \beta =12N/m^2 \end{gathered}$$

Initial Kinetic energy is given by

${\mathrm{K}}_1 = \frac{1}{2}{\mathrm{mu}}^2 = \frac{1}{2}\times 0.1\times 20\times 20 =20 \mathrm{J}$

At the maximum height, the vertical component of the velocity becomes zero, and or horizontal component remains same as $\mathrm{v} = \mathrm{u} \cos 60 = 20\times \frac{1}{2}=10 \mathrm{m}/\mathrm{s}$

So, kinetic energy at the maximum height is

$$\begin{gathered} {\mathrm{K}}_2 = \frac{1}{2}\times 0.1\times 10\times 10 = 5 \mathrm{J} \\ \mathrm{Change} \mathrm{in} \mathrm{KE}, {\mathrm{K}}_1 - {\mathrm{K}}_2 = 20 - 5 = 15 \mathrm{J} \end{gathered}$$

The work done by the force is given by:

where $d\vec{r} = dx \hat{i} + dy \hat{j}$​. Along $x + y = 10$, we use $y = 10 - x$ and $dy = -dx$, resulting in:

Integrating from $x = 0$ to $x = 4$:

After solving, we find:

Thus, the work done by the force is $-112 \, \text{J}$

The work done by the force is given by:

where $d\vec{r} = dx \hat{i} + dy \hat{j}$​. Along $x + y = 10$, we use $y = 10 - x$ and $dy = -dx$, resulting in:

Integrating from $x = 0$ to $x = 4$:

After solving, we find:

Thus, the work done by the force is $-112 \, \text{J}$

Work done by a force is given by the dot product of force and displacement:

$$\begin{gathered} (2\hat{i}+b\hat{j}+\hat{k})·(\hat{i}-2\hat{j}-\hat{k})=0 \\ 2-2b-1=0 \\ b=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{W} = \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}} \\ 0 = \left(\widehat{2\mathrm{i}}+\mathrm{b}\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right).\left(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right) \\ 0 = 2 + \mathrm{b} + 1 \\ \mathrm{b} = - 3 \end{gathered}$$

• Force F = (-2î + 3ĵ) N
• Displacement x = (3î) m
• Work W = F ⋅ x = (–2î + 3ĵ) ⋅ (3î) = (–2)(3) + (3)(0) = –6 J.
• Initial KE (K_i) = 10 J.
• W = K_f – K_i $\Rightarrow$ –6 J = K_f – 10 J $\Rightarrow$ K_f = 10 J – 6 J = 4 J.

(NEW NCERT 11^th Page No. 73)

$$\begin{gathered} \mathrm{W} = \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{d}} \\ 0 = \left(\widehat{2\mathrm{i}}+\mathrm{b}\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right).\left(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\right) \\ 0 = 2 + \mathrm{b} + 1 \\ \mathrm{b} = - 3 \end{gathered}$$