Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 5 m) = 10 m/s.

Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.

Work done by gravity = m × g × h = 2 kg × 9.8 m/s² × 5 m = 98 J.

The net force is equal to the weight of the object: F = mg = 2 kg * 9.8 m/s² = 19.6 N.

Kinetic energy = 0.5 × m × v² = 0.5 × 2 kg × (3 m/s)² = 9 J.

Total mechanical energy = kinetic energy + potential energy. KE = 0.5mv² = 0.5 * 2 * (3)² = 9 J.

Momentum = mass × velocity = 2 kg × 3 m/s = 6 kg·m/s.

Using conservation of momentum and kinetic energy, the final velocity of the 2 kg object after an elastic collision is 1 m/s.

Using F = ma, we find a = F/m = 8 N / 2 kg = 4 m/s².

Potential Energy (PE) = m * g * h = 2 kg * 9.8 m/s^2 * 5 m = 98 J

Momentum p = mv = 2 kg * 4 m/s = 8 kg·m/s.

Momentum p = mv = 2 kg * 5 m/s = 10 kg·m/s.

Using energy conservation: KE_initial = PE_max; 0.5 × 2 kg × (10 m/s)² = 2 kg × 9.8 m/s² × h; h = 5.1 m.

Using the formula v² = u² - 2gh, where v = 0 at max height: 0 = (10)² - 2 * 9.8 * h, h = 5.1 m.

Using energy conservation: KE at bottom = PE at top; 1/2 mv^2 = mgh; h = v^2/(2g) = (10^2)/(2*10) = 5 m

The force acting on the object is its weight: F = mg = 20 kg * 10 m/s² = 200 N.

Net force = applied force - friction = 100 N - 40 N = 60 N. Acceleration a = F/m = 60 N / 20 kg = 3 m/s².

Energy consumed in kilowatt-hours is calculated as E = P * t. Here, P = 2000 W = 2 kW and t = 3 hours. Thus, E = 2 kW * 3 h = 6 kWh.

Using F = ma, we have a = F/m = 100 N / 25 kg = 4 m/s².

Net force = 100 N - 25 N = 75 N. Acceleration a = F/m = 75 N / 25 kg = 3 m/s².

Using F = ma, we find a = F/m = 50 N / 25 kg = 2 m/s².

Using conservation of energy, KE at the bottom = PE at the top. 0.5mv² = mgh. Solving gives h = v²/(2g) = (10)²/(2*9.8) = 5.1 m.

Acceleration a = F/m = 12 N / 3 kg = 4 m/s².

Net force = applied force - friction = 15 N - 5 N = 10 N. Acceleration a = F/m = 10 N / 3 kg = 3.33 m/s².

Net force = applied force - friction = 12 N - 3 N = 9 N. Acceleration a = F/m = 9 N / 3 kg = 3 m/s².

The acceleration down the incline is given by a = g * sin(θ) = 9.8 m/s² * sin(30°) = 4.9 m/s².

Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.

The maximum force exerted by the spring is F = kx. Assuming maximum stretch x = 1 m, F = 200 N/m * 1 m = 200 N.