This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Identify the given values and define variables. Mass of the box, m = 40 kg Initial distance of the box from the open end of the truck, d = 5 m Coefficient of friction between the box and the truck surface, = 0.15 Acceleration of the truck, a_T = 2 m/s^2 We will use the acceleration due to gravity, g = 10 m/s^2. Step 2: Calculate the maximum friction force acting on the box. The normal force (N) acting on the box is equal to its weight, mg. N = m × g N = 40 kg × 10 m/s^2 = 400 N The maximum static friction force (f_friction) is given by N. Once the box starts slipping, this becomes the kinetic friction force. f_friction = × N f_friction = 0.15 × 400 N = 60 N Step 3: Calculate the acceleration of the box. The friction force is the only horizontal force acting on the box. Using Newton's second law (F = ma), the acceleration of the box (a_box) is: a_box = f_frictionm a_box = 60 N40 kg = 1.5 m/s^2 Step 4: Determine the relative acceleration between the truck and the box. The truck accelerates at a_T = 2 m/s^2, and the box accelerates at a_box = 1.5 m/s^2. Since a_T > a_box, the truck moves forward faster than the box, causing the box to slip backward relative to the truck. The relative acceleration (a_rel) of the truck with respect to the box (or the box slipping backward relative to the truck) is: a_rel = a_T - a_box a_rel = 2 m/s^2 - 1.5 m/s^2 = 0.5 m/s^2 Step 5: Calculate the time it takes for the box to fall off the truck. The box falls off when it has moved a relative distance of d = 5 m with respect to the truck. Both the truck and the box start from rest, so the initial relative velocity is 0. Using the kinematic equation s = ut + (1)/(2)at^2: d = 0 × t + (1)/(2) a_rel t^2 5 m = (1)/(2) (0.5 m/s^2) t^2 5 = 0.25 t^2 t^2 = (5)/(0.25) = 20 t = sqrt(20) s Step 6: Calculate the distance traveled by the truck from the starting point. The question asks for the distance from the starting point where the box falls off. This is the distance traveled by the truck during the time t. Using the kinematic equation for the truck: S_T = U_T t + (1)/(2)a_T t^2. Since the truck starts from rest, U_T = 0. S_T = 0 × t + (1)/(2) a_T t^2 S_T = (1)/(2) (2 m/s^2) (20 s^2) S_T = 1 × 20 m S_T = 20 m The distance from the starting point where the box falls off the truck is 20 meters. This corresponds to option (a). The final answer is (a).