Laws of Motion Chapter-Wise Test 8

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

For \( 4 \, \text{kg} \): \( T - f_k - 5 = 4a \), \( f_k = 0.2 \times 4 \times 10 = 8 \, \text{N} \).

\( T - 8 - 5 = 4a \Rightarrow T - 13 = 4a \).

Solve: \( 60 - T = 6a \), \( T - 13 = 4a \Rightarrow 60 - (4a + 13) = 6a \).

\( 60 - 13 - 4a = 6a \Rightarrow 47 = 10a \Rightarrow a = 4.7 \, \text{m/s}^2 \).

\( T - 13 = 4 \times 4.7 \Rightarrow T = 13 + 18.8 = 31.8 \, \text{N} \).

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

\( \mu_s + \tan\theta = 0.3 + 0.364 = 0.664 \), \( 1 - 0.3 \times 0.364 = 1 - 0.1092 = 0.8908 \).

\( v_{\text{max}}^2 = 60 \times 10 \times \frac{0.664}{0.8908} \approx 600 \times 0.7455 \approx 447.3 \).

\( v_{\text{max}} \approx \sqrt{447.3} \approx 21.15 \, \text{m/s} \).

Tangential force causes tangential acceleration: \( F_t = ma_t \).

At max speed, radial forces are balanced, and \( F_t = 2400 \, \text{N} \) acts tangentially.

\( a_t = \frac{F_t}{m} = \frac{2400}{1200} = 2 \, \text{m/s}^2 \).

This remains constant as mass and force are fixed, independent of radial speed limit.

Assume \( 7 \, \text{kg} \) moves down, pulling \( 4 \, \text{kg} \) up.

For \( 7 \, \text{kg} \): \( mg \sin 60^\circ - f_k - T = 7a \).

\( N = mg \cos 60^\circ = 7 \times 10 \times 0.5 = 35 \, \text{N} \).

Friction: \( f_k = 0.2 \times 35 = 7 \, \text{N} \).

\( mg \sin 60^\circ = 70 \times 0.866 = 60.62 \, \text{N} \).

Net force: \( 60.62 - 7 - T = 7a \Rightarrow 53.62 - T = 7a \).

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

Solve: \( 53.62 - T = 7a \), \( T - 40 = 4a \).

Substitute: \( 53.62 - (4a + 40) = 7a \Rightarrow 53.62 - 40 - 4a = 7a \Rightarrow 13.62 = 11a \).

\( a \approx 1.24 \, \text{m/s}^2 \), \( T - 40 = 4 \times 1.24 \Rightarrow T - 40 \approx 4.96 \Rightarrow T \approx 44.96 \, \text{N} \).

Perpendicular component (x-axis) reverses upon rebound.

Initial: \( p_{x1} = m v \cos 37^\circ = 0.18 \times 20 \times 0.8 = 2.88 \, \text{kg m/s} \).

Final: \( p_{x2} = -m v \cos 37^\circ = -0.18 \times 20 \times 0.8 = -2.88 \, \text{kg m/s} \).

Impulse: \( \Delta p_x = -2.88 - 2.88 = -5.76 \, \text{N s} \).

Magnitude: \( |\Delta p_x| = 5.76 \, \text{N s} \approx 5.8 \, \text{N s} \).

Acceleration \( a = \frac{F}{m} = \frac{-20}{4} = -5 \, \text{m/s}^2 \) (negative as it’s retarding).

Use \( v = u + at \), where \( v = 0 \), \( u = 10 \, \text{m/s} \).

Substitute: \( 0 = 10 + (-5)t \).

Solve: \( -10 = -5t \Rightarrow t = \frac{10}{5} = 2 \, \text{s} \).

For \( 10 \, \text{kg} \): \( 10g + 20 - T = 10a \Rightarrow 100 + 20 - T = 10a \Rightarrow 120 - T = 10a \).

For \( 12 \, \text{kg} \): \( T - f_k - F \cos 53^\circ = 12a \).

Normal: \( N = mg + F \sin 53^\circ = 12 \times 10 + 35 \times 0.8 = 120 + 28 = 148 \, \text{N} \).

Friction: \( f_k = 0.3 \times 148 = 44.4 \, \text{N} \).

\( F \cos 53^\circ = 35 \times 0.6 = 21 \, \text{N} \).

Net force: \( T - 44.4 - 21 = 12a \Rightarrow T - 65.4 = 12a \).

Solve: \( 120 - T = 10a \), \( T - 65.4 = 12a \).

Substitute: \( 120 - (12a + 65.4) = 10a \Rightarrow 120 - 65.4 - 12a = 10a \Rightarrow 54.6 = 22a \).

\( a = \frac{54.6}{22} \approx 2.48 \, \text{m/s}^2 \).

The block is on the verge of moving upward, so the net force along the incline is zero, and friction acts downward.

Along incline: \( F \cos (37^\circ - 30^\circ) - mg \sin 37^\circ - f_s = 0 \), where \( \theta = 7^\circ \).

Approximate: \( \cos 7^\circ \approx 0.992 \), \( \sin 7^\circ \approx 0.122 \) (for simplicity, use exact if needed).

Normal force: \( N = mg \cos 37^\circ - F \sin (37^\circ - 30^\circ) \).

\( mg \sin 37^\circ = 5 \times 10 \times 0.6 = 30 \, \text{N} \), \( mg \cos 37^\circ = 5 \times 10 \times 0.8 = 40 \, \text{N} \).

\( N = 40 - F \times 0.122 \), \( f_s = 0.4 (40 - 0.122F) \).

Substitute: \( F \times 0.992 - 30 - 0.4 (40 - 0.122F) = 0 \).

\( 0.992F - 30 - 16 + 0.0488F = 0 \Rightarrow 1.0408F - 46 = 0 \).

\( 1.0408F = 46 \Rightarrow F \approx 44.2 \, \text{N} \).

For \( 5 \, \text{kg} \): \( 5g - T = 5a \Rightarrow 50 - T = 5a \).

For \( 8 \, \text{kg} \): \( T - f_k = 8a \).

Normal: \( N = mg = 8 \times 10 = 80 \, \text{N} \).

Friction: \( f_k = 0.3 \times 80 = 24 \, \text{N} \).

Net force: \( T - 24 = 8a \).

Solve: \( 50 - T = 5a \), \( T - 24 = 8a \).

Substitute: \( 50 - (8a + 24) = 5a \Rightarrow 50 - 24 - 8a = 5a \Rightarrow 26 = 13a \).

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

\( T - 24 = 8 \times 2 \Rightarrow T - 24 = 16 \Rightarrow T = 40 \, \text{N} \).

Net force: \( F_{\text{net}} = mg \sin\theta - f_k + F = ma \).

\( f_k = \mu_k N = 0.3 \times 2 \times 10 \times 0.8 = 4.8 \, \text{N} \).

\( mg \sin\theta = 2 \times 10 \times 0.6 = 12 \, \text{N} \).

\( 12 - 4.8 + F = 2 \times 2 \Rightarrow 7.2 + F = 4 \).

\( F = 4 - 7.2 = -3.2 \, \text{N} \) (up incline), magnitude = \( 3.2 \, \text{N} \).

Impulse along the wall is the change in momentum parallel to the wall (y-axis).

Initial: \( p_{y1} = m v \sin 45^\circ = 0.4 \times 50 \times 0.707 = 14.14 \, \text{kg m/s} \).

Final: \( p_{y2} = m v \sin 60^\circ = 0.4 \times 40 \times 0.866 = 13.856 \, \text{kg m/s} \) (same direction assumed).

Change: \( \Delta p_y = 13.856 - 14.14 = -0.284 \, \text{kg m/s} \).

Magnitude: \( |\Delta p_y| \approx 0.28 \, \text{N s} \).

Verify direction: If rebound angle is opposite, \( p_{y2} = -13.856 \), then \( \Delta p_y = -13.856 - 14.14 = -28 \, \text{N s} \), but question implies along wall magnitude.