Laws of Motion Chapter-Wise Test 7

At bottom: \( T_b - mg = m v_b^2 / r \Rightarrow 25 - 0.5 \times 10 = 0.5 \times v_b^2 / 1.5 \).

\( 25 - 5 = 0.333 v_b^2 \Rightarrow 20 = 0.333 v_b^2 \Rightarrow v_b^2 \approx 60 \Rightarrow v_b \approx 7.75 \, \text{m/s} \).

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

\( 0.5 \times 0.5 \times 60 = 0.5 \times 0.5 \times v_t^2 + 2 \times 0.5 \times 10 \).

\( 15 = 0.25 v_t^2 + 10 \Rightarrow 0.25 v_t^2 = 5 \Rightarrow v_t^2 = 20 \Rightarrow v_t \approx 4.47 \, \text{m/s} \).

At top: \( T_t + mg = m v_t^2 / r \Rightarrow T_t + 5 = 0.5 \times 20 / 1.5 \Rightarrow T_t + 5 \approx 6.67 \Rightarrow T_t \approx 1.67 \, \text{N} \).

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 up incline).

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

Calculate: \( mg = 3 \times 10 = 30 \, \text{N} \), \( mg \sin 53^\circ = 30 \times 0.8 = 24 \, \text{N} \).

\( mg \cos 53^\circ = 30 \times 0.6 = 18 \, \text{N} \), so \( N = 18 + F \times 0.8 \).

Friction: \( f_s = \mu_s N = 0.5 (18 + 0.8F) \).

Substitute: \( F \times 0.6 - 24 - 0.5 (18 + 0.8F) = 0 \).

Expand: \( 0.6F - 24 - 9 - 0.4F = 0 \Rightarrow 0.2F - 33 = 0 \).

Solve: \( 0.2F = 33 \Rightarrow F = \frac{33}{0.2} = 165 \, \text{N} \).

The \( 4.5 \, \text{kg} \) mass descends, pulling the \( 2.5 \, \text{kg} \) block against friction and the opposing force.

For \( 4.5 \, \text{kg} \): \( 4.5g - T = 4.5a \Rightarrow 45 - T = 4.5a \).

For \( 2.5 \, \text{kg} \): \( T - f_k - 10 = 2.5a \).

Normal: \( N = mg = 2.5 \times 10 = 25 \, \text{N} \).

Friction: \( f_k = 0.3 \times 25 = 7.5 \, \text{N} \).

Net force: \( T - 7.5 - 10 = 2.5a \Rightarrow T - 17.5 = 2.5a \).

Solve: \( 45 - T = 4.5a \), \( T - 17.5 = 2.5a \).

Substitute: \( 45 - (2.5a + 17.5) = 4.5a \Rightarrow 45 - 17.5 - 2.5a = 4.5a \Rightarrow 27.5 = 7a \).

\( a = \frac{27.5}{7} \approx 3.93 \, \text{m/s}^2 \).

Initial momentum: \( p_{x1} = 0.27 \times 24 \times 0.6 = 3.888 \, \text{kg m/s} \), \( p_{y1} = 0.27 \times 24 \times 0.8 = 5.184 \, \text{kg m/s} \).

Final momentum: \( p_{x2} = -0.27 \times 24 \times 0.8 = -5.184 \, \text{kg m/s} \), \( p_{y2} = 0.27 \times 24 \times 0.6 = 3.888 \, \text{kg m/s} \).

Change: \( \Delta p_x = -5.184 - 3.888 = -9.072 \, \text{kg m/s} \), \( \Delta p_y = 3.888 - 5.184 = -1.296 \, \text{kg m/s} \).

Magnitude: \( |\Delta p| = \sqrt{(-9.072)^2 + (-1.296)^2} = \sqrt{82.3 + 1.679} \approx \sqrt{83.979} \approx 9.16 \, \text{N s} \).

For \( 2 \, \text{kg} \): \( 2g - T = 2a \Rightarrow 20 - T = 2a \).

For \( 4 \, \text{kg} \): \( T - f_k - 10 = 4a \), \( f_k = 0.25 \times 4 \times 10 = 10 \, \text{N} \).

\( T - 10 - 10 = 4a \Rightarrow T - 20 = 4a \).

Solve: \( 20 - T = 2a \), \( T - 20 = 4a \).

Substitute \( T = 20 - 2a \) into \( 20 - 2a - 20 = 4a \Rightarrow -2a = 4a - 20 \).

\( 6a = 20 \Rightarrow a = \frac{20}{6} \approx 3.33 \, \text{m/s}^2 \).

Perpendicular (x-axis): inicial \( p_x = m v \cos 60^\circ = 0.15 \times 30 \times 0.5 = 2.25 \, \text{kg m/s} \).

Final \( p_x = -2.25 \, \text{kg m/s} \) (rebounds symmetrically).

Impulse \( \Delta p_x = -2.25 - 2.25 = -4.5 \, \text{N s} \).

Magnitude = \( 4.5 \, \text{N s} \).

The \( 4 \, \text{kg} \) mass descends, pulling the \( 3 \, \text{kg} \) block horizontally.

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

For \( 3 \, \text{kg} \): \( T + 15 - f_k = 3a \).

Normal force: \( N = mg = 3 \times 10 = 30 \, \text{N} \).

Friction: \( f_k = 0.2 \times 30 = 6 \, \text{N} \).

Net force: \( T + 15 - 6 = 3a \Rightarrow T + 9 = 3a \).

Solve: \( 40 - T = 4a \), \( T + 9 = 3a \).

Substitute: \( 40 - (3a - 9) = 4a \Rightarrow 40 + 9 - 3a = 4a \Rightarrow 49 = 7a \).

\( a = \frac{49}{7} = 7 \, \text{m/s}^2 \).

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

Along incline: \( F \cos 60^\circ - mg \sin 60^\circ - f_s = 0 \) (horizontal force component opposes gravity and friction).

Normal force: \( N = mg \cos 60^\circ + F \sin 60^\circ \) (horizontal force increases normal force).

Calculate: \( mg = 3.4 \times 10 = 34 \, \text{N} \), \( mg \sin 60^\circ = 34 \times 0.866 = 29.444 \, \text{N} \).

\( mg \cos 60^\circ = 34 \times 0.5 = 17 \, \text{N} \), so \( N = 17 + F \times 0.866 \).

Friction: \( f_s = \mu_s N = 0.4 (17 + 0.866F) \).

Substitute: \( F \times 0.5 - 29.444 - 0.4 (17 + 0.866F) = 0 \).

Expand: \( 0.5F - 29.444 - 6.8 - 0.3464F = 0 \Rightarrow 0.1536F - 36.244 = 0 \).

Solve: \( 0.1536F = 36.244 \Rightarrow F \approx \frac{36.244}{0.1536} \approx 235.94 \, \text{N} \).

The scale reads the normal force.

Net force: \( N - mg = ma \), where \( a = -2 \, \text{m/s}^2 \).

\( N - 60 \times 10 = 60 \times (-2) \).

\( N - 600 = -120 \Rightarrow N = 480 \, \text{N} \).

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

For \( 7 \, \text{kg} \): \( T + 10 - f_k - 20 = 7a \), \( f_k = 0.4 \times 7 \times 10 = 28 \, \text{N} \).

\( T + 10 - 28 - 20 = 7a \Rightarrow T - 38 = 7a \).

Solve: \( 50 - T = 5a \), \( T - 38 = 7a \Rightarrow 50 - (7a + 38) = 5a \Rightarrow 12 = 12a \Rightarrow a = 1 \, \text{m/s}^2 \).

\( T - 38 = 7 \times 1 \Rightarrow T = 45 \, \text{N} \).