Laws Of Motion Chapter-Wise Test 1

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 0.45 \times 10 = 0.45 \times (14)^2 / 1.8 \).

\( T_b - 4.5 = 0.45 \times 108.89 \Rightarrow T_b - 4.5 \approx 49 \Rightarrow T_b \approx 53.5 \, \text{N} \).

Energy: \( \frac{1}{2} m v_b^2 = \frac{1}{2} m v_t^2 + 2mg \).

\( 0.5 \times 0.45 \times 196 = 0.5 \times 0.45 \times v_t^2 + 2 \times 0.45 \times 10 \).

\( 44.1 = 0.225 v_t^2 + 9 \Rightarrow 0.225 v_t^2 = 35.1 \Rightarrow v_t^2 \approx 156 \Rightarrow v_t \approx 12.49 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 4.5 = 0.45 \times 156 / 1.8 \Rightarrow T_t + 4.5 \approx 39 \Rightarrow T_t \approx 34.5 \, \text{N} \).

Tension provides centripetal force: \( T = m v^2 / r \).

Substitute: \( 10 = 0.25 \times v^2 / 2 \).

\( 10 = 0.125 v^2 \Rightarrow v^2 = \frac{10}{0.125} = 80 \).

\( v = \sqrt{80} \approx 8.94 \, \text{m/s} \).

Block moves downward, so friction acts upward.

Along incline: \( F \sin 30^\circ + mg \sin 30^\circ - f_s = 0 \).

Normal: \( N = mg \cos 30^\circ + F \cos 30^\circ \).

\( mg \sin 30^\circ = 4 \times 10 \times 0.5 = 20 \, \text{N} \), \( mg \cos 30^\circ = 40 \times 0.866 = 34.64 \, \text{N} \).

\( N = 34.64 + F \times 0.866 \), \( f_s = 0.3 (34.64 + 0.866F) \).

Substitute: \( F \times 0.5 + 20 - 0.3 (34.64 + 0.866F) = 0 \).

\( 0.5F + 20 - 10.392 - 0.2598F = 0 \Rightarrow 0.2402F + 9.608 = 0 \).

\( 0.2402F = -9.608 \Rightarrow F \approx -40 \) (impossible, adjust: \( F \sin 30^\circ = f_s - mg \sin 30^\circ \)).

Correct: \( f_s = 0.3 \times 34.64 = 10.392 \), \( F \times 0.5 = 10.392 - 20 \Rightarrow F = -19.2 \) (retry).

\( 0.5F = 20 - 10.392 \Rightarrow 0.5F = 9.608 \Rightarrow F \approx 19.2 \, \text{N} \).

Optimum speed: \( v_0 = \sqrt{rg \tan\theta} \).

Substitute: \( r = 40 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \), \( \tan 20^\circ = 0.364 \).

\( v_0^2 = 40 \times 10 \times 0.364 = 145.6 \).

\( v_0 = \sqrt{145.6} \approx 12.07 \, \text{m/s} \).

For two masses over a pulley, acceleration is \( a = \frac{(m_2 - m_1)g}{m_1 + m_2} \).

Here, \( m_2 = 6 \, \text{kg} \) (heavier mass), \( m_1 = 4 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \).

Numerator: \( (m_2 - m_1)g = (6 - 4) \times 10 = 20 \).

Denominator: \( m_1 + m_2 = 4 + 6 = 10 \).

So, \( a = \frac{20}{10} = 2 \, \text{m/s}^2 \).

The block is about to move upward, so friction acts downward along the incline.

Along incline: \( F \cos 53^\circ - mg \sin 53^\circ - f_s = 0 \) (horizontal force component opposes gravity and friction).

Normal force: \( N = mg \cos 53^\circ + F \sin 53^\circ \) (horizontal force increases normal force).

Calculate: \( mg = 5.2 \times 10 = 52 \, \text{N} \), \( mg \sin 53^\circ = 52 \times 0.8 = 41.6 \, \text{N} \).

\( mg \cos 53^\circ = 52 \times 0.6 = 31.2 \, \text{N} \), so \( N = 31.2 + F \times 0.8 \).

Friction: \( f_s = \mu_s N = 0.3 (31.2 + 0.8F) \).

Substitute: \( F \times 0.6 - 41.6 - 0.3 (31.2 + 0.8F) = 0 \).

Expand: \( 0.6F - 41.6 - 9.36 - 0.24F = 0 \Rightarrow 0.36F - 50.96 = 0 \).

Solve: \( 0.36F = 50.96 \Rightarrow F \approx \frac{50.96}{0.36} \approx 141.56 \, \text{N} \).

For \( 7.8 \, \text{kg} \): \( 7.8g - T = 7.8a \Rightarrow 78 - T = 7.8a \).

For \( 5.8 \, \text{kg} \): \( T - mg \sin 45^\circ - f_k = 5.8a \).

\( N = mg \cos 45^\circ = 5.8 \times 10 \times 0.707 = 41.006 \, \text{N} \).

Friction: \( f_k = 0.25 \times 41.006 \approx 10.2515 \, \text{N} \).

\( mg \sin 45^\circ = 58 \times 0.707 = 41.006 \, \text{N} \).

Net force: \( T - 41.006 - 10.2515 = 5.8a \Rightarrow T - 51.2575 = 5.8a \).

Solve: \( 78 - T = 7.8a \), \( T - 51.2575 = 5.8a \).

Substitute: \( 78 - (5.8a + 51.2575) = 7.8a \Rightarrow 78 - 51.2575 - 5.8a = 7.8a \Rightarrow 26.7425 = 13.6a \).

\( a \approx 1.97 \, \text{m/s}^2 \), \( T - 51.2575 = 5.8 \times 1.97 \Rightarrow T - 51.2575 \approx 11.426 \Rightarrow T \approx 62.68 \, \text{N} \).

Angular speed: \( \omega = 54 \times \frac{2\pi}{60} = \frac{9\pi}{5} \, \text{rad/s} \).

\( \omega^2 = \left(\frac{9\pi}{5}\right)^2 = \frac{81\pi^2}{25} \approx 31.99 \).

Tension: \( T = m \omega^2 r = 0.31 \times 31.99 \times 1.3 \).

\( T \approx 0.31 \times 31.99 \times 1.3 \approx 12.9 \, \text{N} \).

Scale reads normal force.

Net force: \( N - mg = ma \).

\( N - 80 \times 10 = 80 \times 1.5 \).

\( N - 800 = 120 \Rightarrow N = 920 \, \text{N} \).

Angular speed: \( \omega = 45 \times \frac{2\pi}{60} = \frac{3\pi}{2} \, \text{rad/s} \).

\( \omega^2 = \left(\frac{3\pi}{2}\right)^2 = \frac{9\pi^2}{4} \approx 22.21 \).

Tension: \( T = m \omega^2 r = 0.3 \times 22.21 \times 1.6 \).

\( T \approx 0.3 \times 22.21 \times 1.6 \approx 10.66 \, \text{N} \).