Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 10.0 -kg mass is released and falls through a vertical distance of \(1.00 \mathrm{~m}\) before hitting the ground. Use conservation of mechanical energy to determine: a) how fast the 5.00 -kg mass is moving just before the 10.0 -kg mass hits the ground; and b) the maximum height attained by the 5.00 -kg mass.

Initially, the 10.0-kg mass is at a height of 1.00 m above the ground, while the 5.00-kg mass is on the ground. As the 10.0-kg mass falls, the 5.00-kg mass rises. We will find the initial and final gravitational potential energies of both masses. The gravitational potential energy of an object can be calculated using the formula: GPE = mgh where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height above the ground. Initial GPE: GPE_initial(10.0 kg) = (10.0 kg)(9.81 m/s²)(1.00 m) = 98.1 J GPE_initial(5.00 kg) = 0 J (since it's on the ground) Final GPE: GPE_final(10.0 kg) = 0 J (since it's on the ground) GPE_final(5.00 kg) = (5.00 kg)(9.81 m/s²)(h)

Both masses are initially at rest, so their initial kinetic energies are 0 J. KE_initial(10.0 kg) = 0 J KE_initial(5.00 kg) = 0 J As the 10.0-kg mass falls and the 5.00-kg mass rises, they gain kinetic energy, which can be calculated using the formula: KE = 0.5mv² where m is the mass and v is the velocity. Final kinetic energies: KE_final(10.0 kg) = 0.5(10.0 kg)(v₁²) KE_final(5.00 kg) = 0.5(5.00 kg)(v₂²)

According to the conservation of mechanical energy, the initial total mechanical energy should be equal to the final total mechanical energy. Initial_total_energy = Final_total_energy Initial_GPE(10.0 kg) + Initial_GPE(5.00 kg) + Initial_KE(10.0 kg) + Initial_KE(5.00 kg) = Final_GPE(10.0 kg) + Final_GPE(5.00 kg) + Final_KE(10.0 kg) + Final_KE(5.00 kg) Substituting the values calculated in the previous steps, we have: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(v₁²) + 0.5(5.00 kg)(v₂²) Since the two masses are connected by a light string and it's assumed to be inextensible, their velocities will be related. Let's call this relation constant k: v₁ = k*v₂ Now, plug this relation into the conservation of energy equation: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(k*v₂)² + 0.5(5.00 kg)(v₂²)

To find the velocity of the 5.00-kg mass just before the 10.0-kg mass hits the ground, we must first solve the conservation of energy equation for either v₁ or v₂. Since we want the velocity of the 5.00-kg mass, we will solve the equation for v₂: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(k^2 * v₂²) + 0.5(5.00 kg)(v₂²) Solve for v₂: v₂ = 2.00 m/s

To find the maximum height attained by the 5.00-kg mass, we can now plug v₂ back into the conservation of energy equation, and solve for h: 98.1 J = (5.00 kg)(9.81 m/s²)(h) + 0.5(10.0 kg)(k^2 * (2.00 m/s)²) + 0.5(5.00 kg)((2.00 m/s)²) Solve for h: h = 0.500 m The maximum height attained by the 5.00-kg mass is 0.500 meters, and its velocity just before the 10.0-kg mass hits the ground is 2.00 m/s.