A \(100 \%\) -efficient electric motor is lifting a 15 -N weight at \(25 \mathrm{cm} / \mathrm{s} .\) How much current does it draw from a \(6.0-\mathrm{V}\) battery?

From the given values in the exercise, one can calculate the power the motor needs. The motor uses power to lift the weight at a certain speed; in physics, power is defined as the rate at which work is done or energy is transferred. In this case, the motor does work by lifting the weight against gravity. The amount of work done per unit time (which is the same as power) equals the weight of the object times the velocity at which the object is lifted. Here, we identify that a weight of 15 N is lifted at a speed of 25 cm/s or 0.25 m/s (since we express all quantities here in the base SI units). The power required by the motor \(P\) can be calculated using the formula: \[P = w \cdot v \] where: \(P\) is the power, \(w\) is the weight, \(v\) is the speed.

The problem states, that the electric motor is 100% efficient at converting electrical power from a battery into mechanical power for lifting the weight. This allows us to infer that the electrical power provided by the battery is equal to the mechanical power required to lift the weight. Power in circuits is also given by \(P=IV\), where \(I\) is the current and \(V\) is the voltage. We know that \(V = 6.0 V\) and we can solve the power equation for \(I\): \(I = P/V\)