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.
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Here is the solution to your problem: 5.1 State the work-energy theorem in words. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. 5.2 Using ENERGY PRINCIPLES ONLY, calculate the kinetic energy of the crate at point B. The crate starts at rest at point A, so its initial kinetic energy K_A = 0 J. A constant horizontal force F = 12 N acts on the crate over a distance d_AB = 7 m. The surface from A to B is frictionless. The work done by the force F is given by: W_F = F · d_AB According to the work-energy theorem, the net work done on the crate is equal to the change in its kinetic energy: W_net = K = K_B - K_A Since K_A = 0 J and the only force doing work is F: K_B = W_F K_B = (12 N)(7 m) K_B = 84 J The kinetic energy of the crate at point B is 84 J. 5.3 Draw a free-body diagram showing ALL the forces acting on the crate as it moves down the incline. As the crate moves down the incline, the forces acting on it are: • Gravitational force (Weight): Acts vertically downwards from the center of mass of the crate. • Normal force: Acts perpendicular to the inclined surface, pointing away from the surface. • Kinetic frictional force: Acts parallel to the inclined surface, pointing upwards along the incline (opposite to the direction of motion). 5.4 Using ENERGY PRINCIPLES ONLY, determine whether the crate will pass point C. From point B to C, the crate moves down an incline. Initial kinetic energy at B: K_B = 84 J Mass of the crate: m = 3 kg Distance from B to C: d_BC = 6.8 m Angle of inclination: = 20^ Constant frictional force: f_k = 21 N Acceleration due to gravity: g = 9.8 m/s^2 We use the work-energy theorem from B to C: W_net, BC = K_C - K_B. The forces doing work are gravity and friction. Step 1: Calculate the work done by gravity (W_g). The component of gravity parallel to the incline is mg . This component acts in the direction of motion. W_g = (mg ) d_BC W_g = (3 kg)(9.8 m/s^2) (20^) (6.8 m) W_g = (29.4 N)(0.3420)(6.8 m) W_g ≈ 68.45 J Step 2: Calculate the work done by friction (W_f). Friction opposes the motion, so the work done by friction is negative. W_f = -f_k · d_BC W_f = -(21 N)(6.8 m) W_f = -142.8 J Step 3: Calculate the net work done from B to C (W_net, BC). W_net, BC = W_g + W_f W_net, BC = 68.45 J - 142.8 J W_net, BC = -74.35 J Step 4: Calculate the kinetic energy at C (K_C). Using the work-energy theorem: K_C = K_B + W_net, BC K_C = 84 J + (-74.35 J) K_C = 9.65 J Since the kinetic energy at point C (K_C = 9.65 J) is positive, the crate still has kinetic energy when it reaches point C. Therefore, the crate will pass point C. The crate will pass point C.