1. **State the problem:** Calculate the magnitude of the applied force pushing a 23 kg luggage up a ramp inclined at 15° with kinetic friction coefficient 0.18, given it accelerates from rest to 3.7 m/s over 1.1 m. 2. **Known values:** - Mass $m = 23$ kg - Initial velocity $v_0 = 0$ m/s - Final velocity $v_f = 3.7$ m/s - Distance along ramp $d = 1.1$ m - Angle of incline $\theta = 15^\circ$ - Coefficient of kinetic friction $\mu = 0.18$ - Gravitational acceleration $g = 9.81$ m/s$^2$ 3. **Find acceleration $a$ using kinematic equation:** $$v_f^2 = v_0^2 + 2ad \implies a = \frac{v_f^2 - v_0^2}{2d} = \frac{3.7^2 - 0}{2 \times 1.1} = \frac{13.69}{2.2} = 6.22\text{ m/s}^2$$ 4. **Calculate forces:** - Weight force $F_g = mg = 23 \times 9.81 = 225.63$ N (downward) - Normal force $F_N = F_g \cos \theta = 225.63 \times \cos 15^\circ = 225.63 \times 0.9659 = 217.99$ N - Friction force $F_f = \mu F_N = 0.18 \times 217.99 = 39.24$ N (opposes motion) 5. **Calculate net force along ramp using Newton's second law:** $$F_{net} = ma = 23 \times 6.22 = 143.06\text{ N}$$ 6. **Calculate component of weight along ramp:** $$F_{g,\parallel} = F_g \sin \theta = 225.63 \times \sin 15^\circ = 225.63 \times 0.2588 = 58.43\text{ N}$$ 7. **Calculate applied force $F_a$ parallel to ramp:** Net force is applied force minus friction and weight components: $$F_a - F_f - F_{g,\parallel} = F_{net} \implies F_a = F_{net} + F_f + F_{g,\parallel} = 143.06 + 39.24 + 58.43 = 240.73\text{ N}$$ **Final answer:** The magnitude of the applied force is approximately **240.7 N**.