An electric motor on a forklift truck turns at a speed of 600 revolutions per minute (rpm) for 4 seconds. It produces a torque of 20 N\cdot m to turn a hydraulic pump. The hydraulic system lifts a 154-kg box a vertical distance of 2.5 m. What is the effi

Efficiency of the system can be calculated using the formula: Efficiency = (output power / input power) x 100% where, output power = work done by the hydraulic system to lift the box, input power = power supplied to the electric motor. Here, the output power can be calculated as: Output power = work done / time taken = mgh / t where, m = mass of box = 154 kg g = acceleration due to gravity = 9.8 m/s^2 h = height lifted = 2.5 m t = time taken to lift the box = (600 rpm / 60) s = 10 s Therefore, Output power = 154 x 9.8 x 2.5 / 10 = 379.5 W Now, the input power can be calculated as: Input power = torque x angular velocity = (20 N.m) x (2? x 600 / 60) = 1256.6 W Therefore, the efficiency of the system is: Efficiency = (output power / input power) x 100% = (379.5 / 1256.6) x 100% = 30.2% Hence, the efficiency of the hydraulic system is 30.2%. This means that only 30.2% of the input power supplied to the electric motor is effectively utilized to lift the box. The rest of the power is lost due to various inefficiencies such as friction, heat losses, etc.