Laws of Motion Chapter-Wise Test 10

Net force \( F_{\text{net}} = ma \), where \( a = 4 \, \text{m/s}^2 \).

Thrust \( T \) must overcome weight \( mg \) and provide net force: \( T - mg = ma \).

Substitute: \( T - 5000 \times 10 = 5000 \times 4 \).

\( T - 50000 = 20000 \).

So, \( T = 50000 + 20000 = 70000 \, \text{N} \).

Initial momentum of the nucleus is zero (at rest).

By conservation of momentum: \( m_1 v_1 + m_2 v_2 = 0 \).

Substitute: \( 2 \times 6 + 3 \times v_2 = 0 \).

\( 12 + 3 v_2 = 0 \Rightarrow 3 v_2 = -12 \).

\( v_2 = -4 \, \text{m/s} \), so speed = \( 4 \, \text{m/s} \) (opposite direction).

The block is on the verge of moving up, so friction acts downward along the incline.

Along incline: \( F \cos 30^\circ - mg \sin 30^\circ - f_s = 0 \) (horizontal force component opposes gravity and friction).

Normal force: \( N = mg \cos 30^\circ + F \sin 30^\circ \) (horizontal force increases normal force).

Calculate: \( mg = 2.5 \times 10 = 25 \, \text{N} \), \( mg \sin 30^\circ = 25 \times 0.5 = 12.5 \, \text{N} \).

\( mg \cos 30^\circ = 25 \times 0.866 = 21.65 \, \text{N} \), so \( N = 21.65 + F \times 0.5 \).

Friction: \( f_s = \mu_s N = 0.4 (21.65 + 0.5F) \).

Substitute: \( F \times 0.866 - 12.5 - 0.4 (21.65 + 0.5F) = 0 \).

Expand: \( 0.866F - 12.5 - 8.66 - 0.2F = 0 \Rightarrow 0.666F - 21.16 = 0 \).

Solve: \( 0.666F = 21.16 \Rightarrow F \approx \frac{21.16}{0.666} \approx 31.77 \, \text{N} \).

For \( 6 \, \text{kg} \): \( mg \sin\theta - f_k - T = 6a \), \( f_k = \mu_k N = 0.1 \times 6 \times 10 \times 0.6 = 3.6 \, \text{N} \).

\( 6 \times 10 \times 0.8 - 3.6 - T = 6a \Rightarrow 48 - 3.6 - T = 6a \Rightarrow 44.4 - T = 6a \).

For \( 4 \, \text{kg} \): \( T - 4g = 4a \Rightarrow T - 40 = 4a \).

Solve: \( 44.4 - T = 6a \), \( T - 40 = 4a \).

\( 44.4 - (4a + 40) = 6a \Rightarrow 44.4 - 40 - 4a = 6a \Rightarrow 4.4 = 10a \Rightarrow a = 0.44 \, \text{m/s}^2 \).

For \( 3 \, \text{kg} \): \( 3g - T = 3a \Rightarrow 30 - T = 3a \).

For \( 2 \, \text{kg} \): \( T + 10 - f_k = 2a \).

Normal: \( N = mg = 2 \times 10 = 20 \, \text{N} \).

Friction: \( f_k = 0.25 \times 20 = 5 \, \text{N} \).

Net force: \( T + 10 - 5 = 2a \Rightarrow T + 5 = 2a \).

Solve: \( 30 - T = 3a \), \( T + 5 = 2a \).

Substitute: \( 30 - (2a - 5) = 3a \Rightarrow 30 + 5 - 2a = 3a \Rightarrow 35 = 5a \).

\( a = 7 \, \text{m/s}^2 \).

For the box: Forces are weight downward, man’s force downward, and normal force from floor upward.

Net force: \( N_b - mg - F = ma \Rightarrow N_b - 20 \times 10 - 100 = 20 \times 4 \).

\( N_b - 200 - 100 = 80 \Rightarrow N_b - 300 = 80 \Rightarrow N_b = 380 \, \text{N} \).

For the man: Forces are weight downward, normal force from floor upward, and reaction force from box upward (Newton’s third law).

Net force: \( N_m - mg - (-F) = ma \Rightarrow N_m - 60 \times 10 + 100 = 60 \times 4 \).

\( N_m - 600 + 100 = 240 \Rightarrow N_m - 500 = 240 \Rightarrow N_m = 740 \, \text{N} \).

Total normal force on lift floor: \( N = N_m + N_b = 740 + 380 = 1120 \, \text{N} \).

At the point of impending motion, net force along horizontal = 0 and vertical = 0.

Horizontal: \( T \cos 30^\circ = f_s \), where \( f_s = \mu_s N \).

Vertical: \( N + T \sin 30^\circ = mg \Rightarrow N = mg - T \sin 30^\circ \).

Substitute: \( mg = 5 \times 10 = 50 \, \text{N} \), \( \sin 30^\circ = 0.5 \), \( N = 50 - 0.5T \).

Friction: \( f_s = 0.5 \times (50 - 0.5T) \).

Horizontal: \( T \cos 30^\circ = 0.5 (50 - 0.5T) \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).

\( T \frac{\sqrt{3}}{2} = 0.5 (50 - 0.5T) \Rightarrow T \sqrt{3} = 50 - 0.5T \).

\( T \sqrt{3} + 0.5T = 50 \Rightarrow T (\sqrt{3} + 0.5) = 50 \Rightarrow T = \frac{50}{1.732 + 0.5} \approx 22.47 \, \text{N} \).

Truck’s velocity at \( t = 5 \, \text{s} \): \( v = 0 + 2 \times 5 = 10 \, \text{m/s} \).

Mass retains this horizontal velocity when dropped.

No horizontal acceleration after drop (air resistance neglected).

Distance = \( v \times t = 10 \times 2 = 20 \, \text{m} \).

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow T_b - 0.5 \times 10 = 0.5 \times (12)^2 / 3 \).

\( T_b - 5 = 0.5 \times 48 \Rightarrow T_b - 5 = 24 \Rightarrow T_b = 29 \, \text{N} \) (tangential force doesn’t affect radial).

Energy loss: Work by tangential force over half-circle (arc length \( \pi r = \pi \times 3 = 9.42 \, \text{m} \)) reduces kinetic energy.

Work: \( F_t \times s = 2 \times 9.42 = 18.84 \, \text{J} \).

Initial KE: \( \frac{1}{2} m v_b^2 = 0.5 \times 0.5 \times 144 = 36 \, \text{J} \).

PE change: \( 2mg = 2 \times 0.5 \times 10 \times 3 = 30 \, \text{J} \).

Final KE at top: \( 36 - 18.84 - 30 = -12.84 \, \text{J} \) (impossible, adjust for min speed at top).

Minimum \( v_t^2 = 0 \), but use conservation: \( \frac{1}{2} m v_b^2 - F_t \times 2r = \frac{1}{2} m v_t^2 + 2mg \).

\( 36 - 2 \times 6 = 0.5 \times 0.5 \times v_t^2 + 15 \Rightarrow 24 = 0.25 v_t^2 + 15 \Rightarrow 9 = 0.25 v_t^2 \Rightarrow v_t^2 = 36 \Rightarrow v_t = 6 \, \text{m/s} \).

Top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 5 = 0.5 \times 36 / 3 \Rightarrow T_t + 5 = 6 \Rightarrow T_t = 1 \, \text{N} \).

For \( 6 \, \text{kg} \): \( 6g - T = 6a \Rightarrow 60 - T = 6a \).

For \( 9 \, \text{kg} \): \( T + F \cos 30^\circ - f_k = 9a \), \( N = mg - F \sin 30^\circ \).

\( N = 9 \times 10 - 25 \times 0.5 = 90 - 12.5 = 77.5 \, \text{N} \).

\( f_k = 0.2 \times 77.5 = 15.5 \, \text{N} \), \( F \cos 30^\circ = 25 \times 0.866 \approx 21.65 \, \text{N} \).

\( T + 21.65 - 15.5 = 9a \Rightarrow T + 6.15 = 9a \).

Solve: \( 60 - T = 6a \), \( T + 6.15 = 9a \Rightarrow 60 - (9a - 6.15) = 6a \Rightarrow 66.15 = 15a \Rightarrow a \approx 4.41 \, \text{m/s}^2 \).