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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5.1: Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, relative to a reference point. 5.2: Step 1: Identify the given values and the formula for gravitational potential energy. The mass of the hammer is m = 1 kg. The height of point X is h = 35 m. The acceleration due to gravity is g = 9.8 m/s^2. The formula for gravitational potential energy is E_p = mgh. Step 2: Substitute the values into the formula and calculate. E_p = (1 kg)(9.8 m/s^2)(35 m) E_p = 343 J The gravitational potential energy of the hammer at point X is 343 J. 5.3: The principle of conservation of mechanical energy states that in an isolated system where only conservative forces (like gravity) do work, the total mechanical energy (the sum of kinetic and potential energy) remains constant. 5.4: Step 1: Apply the principle of conservation of mechanical energy. Since air resistance is ignored, mechanical energy is conserved. Total mechanical energy at X = Total mechanical energy at Y E_mech, X = E_mech, Y E_p, X + E_k, X = E_p, Y + E_k, Y Step 2: Determine the energy components at point X. At point X, the hammer is dropped, so its initial velocity is v_X = 0 m/s. The height is h_X = 35 m. E_p, X = mgh_X = (1 kg)(9.8 m/s^2)(35 m) = 343 J E_k, X = (1)/(2)mv_X^2 = (1)/(2)(1 kg)(0 m/s)^2 = 0 J So, E_mech, X = 343 J + 0 J = 343 J. Step 3: Determine the energy components at point Y. At point Y, the hammer is on the ground, so its height is h_Y = 0 m. Let the speed at point Y be v_Y. E_p, Y = mgh_Y = (1 kg)(9.8 m/s^2)(0 m) = 0 J E_k, Y = (1)/(2)mv_Y^2 = (1)/(2)(1 kg)v_Y^2 Step 4: Equate the total mechanical energies and solve for v_Y. 343 J = 0 J + (1)/(2)(1 kg)v_Y^2 343 = (1)/(2)v_Y^2 686 = v_Y^2 v_Y = sqrt(686) v_Y ≈ 26.19 m/s The speed of the hammer at the instant it strikes the ground at point Y is 26.19 m/s. Send me the next one 📸