Laws of Motion Chapter-Wise Test 9

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 0.42 \times 10 = 0.42 \times (12)^2 / 1.4 \).

\( T_b - 4.2 = 0.42 \times 102.86 \Rightarrow T_b - 4.2 \approx 43.2 \Rightarrow T_b \approx 47.4 \, \text{N} \).

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

\( 0.5 \times 0.42 \times 144 = 0.5 \times 0.42 \times v_t^2 + 2 \times 0.42 \times 10 \).

\( 30.24 = 0.21 v_t^2 + 8.4 \Rightarrow 0.21 v_t^2 = 21.84 \Rightarrow v_t^2 \approx 104 \Rightarrow v_t \approx 10.2 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 4.2 = 0.42 \times 104 / 1.4 \Rightarrow T_t + 4.2 \approx 31.2 \Rightarrow T_t \approx 27 \, \text{N} \).

The block is in equilibrium, so the net force is zero.

The weight of the block acts downward: \( mg = 5 \times 10 = 50 \, \text{N} \).

The tension \( T \) in the rope acts upward and balances the weight.

Thus, \( T = mg = 50 \, \text{N} \).

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

Substitute: \( 15 = 0.4 \times v^2 / 1.8 \).

\( 15 = 0.222 v^2 \Rightarrow v^2 = \frac{15}{0.222} \approx 67.57 \).

\( v = \sqrt{67.57} \approx 8.22 \, \text{m/s} \).

Maximum speed: \( v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} \).

Numerator: \( \mu_s + \tan\theta = 0.2 + 0.466 = 0.666 \).

Denominator: \( 1 - 0.2 \times 0.466 = 1 - 0.0932 = 0.9068 \).

\( v_{\text{max}}^2 = 40 \times 10 \times \frac{0.666}{0.9068} \approx 400 \times 0.7342 \approx 293.68 \).

\( v_{\text{max}} = \sqrt{293.68} \approx 17.14 \, \text{m/s} \).

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

\( 32000 - 2200 \times 10 = 2200 \times 3.5 \Rightarrow 32000 - 22000 = 7700 \).

Net force from ejection: \( T_{\text{ejection}} = ma = 7700 \, \text{N} \).

Thrust from gas: \( T_{\text{ejection}} = \frac{dm}{dt} \times v_{\text{rel}} \).

\( 7700 = \frac{dm}{dt} \times 80 \Rightarrow \frac{dm}{dt} = \frac{7700}{80} = 96.25 \, \text{kg/s} \).

Optimum speed: \( v_0 = \sqrt{rg \tan\theta} \) (no friction contribution).

Substitute: \( r = 60 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \), \( \tan 20^\circ = 0.364 \).

\( v_0^2 = 60 \times 10 \times 0.364 = 600 \times 0.364 = 218.4 \).

\( v_0 = \sqrt{218.4} \approx 14.78 \, \text{m/s} \).

The system accelerates as the \( 5 \, \text{kg} \) mass moves downward and the \( 4 \, \text{kg} \) block moves horizontally.

For the \( 5 \, \text{kg} \) mass (vertical): Forces are weight downward, tension upward, and opposing force upward.

Net force: \( 5g - T - 10 = 5a \Rightarrow 50 - T - 10 = 5a \Rightarrow 40 - T = 5a \).

For the \( 4 \, \text{kg} \) block (horizontal): Forces are tension, kinetic friction, and the horizontal component of the \( 30 \, \text{N} \) force.

Normal force: \( N = mg - F \sin 53^\circ = 4 \times 10 - 30 \times 0.8 = 40 - 24 = 16 \, \text{N} \).

Friction: \( f_k = \mu_k N = 0.3 \times 16 = 4.8 \, \text{N} \).

Horizontal component of force: \( F \cos 53^\circ = 30 \times 0.6 = 18 \, \text{N} \).

Net force: \( T + 18 - 4.8 = 4a \Rightarrow T + 13.2 = 4a \).

Solve the two equations: \( 40 - T = 5a \) and \( T + 13.2 = 4a \).

Substitute \( T = 40 - 5a \) into the second: \( 40 - 5a + 13.2 = 4a \).

Combine: \( 53.2 = 4a + 5a \Rightarrow 53.2 = 9a \).

Solve: \( a = \frac{53.2}{9} \approx 5.91 \, \text{m/s}^2 \).

For \( 7 \, \text{kg} \): \( 7g - T = 7a \Rightarrow 70 - T = 7a \).

For \( 10 \, \text{kg} \): \( T - f_k - F \cos 37^\circ = 10a \).

Normal force: \( N = mg + F \sin 37^\circ = 10 \times 10 + 40 \times 0.6 = 100 + 24 = 124 \, \text{N} \).

Friction: \( f_k = 0.35 \times 124 = 43.4 \, \text{N} \).

\( F \cos 37^\circ = 40 \times 0.8 = 32 \, \text{N} \).

Net force: \( T - 43.4 - 32 = 10a \Rightarrow T - 75.4 = 10a \).

Solve: \( 70 - T = 7a \), \( T - 75.4 = 10a \).

Substitute: \( 70 - (10a + 75.4) = 7a \Rightarrow 70 - 75.4 - 10a = 7a \Rightarrow -5.4 = 17a \).

\( a = \frac{-5.4}{17} \approx -0.318 \, \text{m/s}^2 \) (opposite direction, \( 7 \, \text{kg} \) ascends).

\( T - 75.4 = 10 \times (-0.318) \Rightarrow T - 75.4 = -3.18 \Rightarrow T \approx 72.22 \, \text{N} \).

For \( 7.5 \, \text{kg} \): \( 7.5g - T = 7.5a \Rightarrow 75 - T = 7.5a \).

For \( 5.5 \, \text{kg} \): \( T - mg \sin 37^\circ - f_k = 5.5a \).

\( N = mg \cos 37^\circ = 5.5 \times 10 \times 0.8 = 44 \, \text{N} \).

Friction: \( f_k = 0.2 \times 44 = 8.8 \, \text{N} \).

\( mg \sin 37^\circ = 55 \times 0.6 = 33 \, \text{N} \).

Net force: \( T - 33 - 8.8 = 5.5a \Rightarrow T - 41.8 = 5.5a \).

Solve: \( 75 - T = 7.5a \), \( T - 41.8 = 5.5a \).

Substitute: \( 75 - (5.5a + 41.8) = 7.5a \Rightarrow 75 - 41.8 - 5.5a = 7.5a \Rightarrow 33.2 = 13a \).

\( a = \frac{33.2}{13} \approx 2.55 \, \text{m/s}^2 \).

Truck’s velocity after \( 5 \, \text{s} \): \( v = u + at = 0 + 2.2 \times 5 = 11 \, \text{m/s} \).

Stone inherits this horizontal velocity upon drop.

No horizontal forces act after drop (air resistance neglected).

Distance: \( s = v \times t = 11 \times 2 = 22 \, \text{m} \).