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When trying to solve problems involving moving objects, Newton's Second Law is a crucial tool. This law states that the force on an object is equal to its mass multiplied by its acceleration. This can be written as:\[ F = m \cdot a \]In simple terms, the law indicates how an object will move if a force is applied to it. Here, we have a mass of \(9.0 \times 10^{-5} \mathrm{kg}\) and an unknown force, which we need to determine to help find the electric field. By understanding the relationship between mass, force, and acceleration, we can proceed to connect other concepts like electric force with this foundational law. Remember, any acceleration that occurs is directly proportional to the force applied, and inversely proportional to the object's mass.

Electric force is a specifically defined force experienced by a charged object in an electric field. According to our formula, the force can also be calculated by the product of the charge and the magnitude of the electric field:\[ F = q \cdot E \]For the exercise in question, the assignment is to determine how the electric force relates directly to the object’s acceleration. The charge on the object is given as \( q = 7.5 \times 10^{-6} \mathrm{C} \), which, along with the electric field strength, defines the electric force. The beauty of the electric force lies in its direct proportionality to both the electric field and the charge, making these problems easier to handle by breaking them into known and direct equations. Once we connect this with the calculated force from Newton's Second Law, solving for the electric field becomes straightforward.

Acceleration is a measure of how quickly the velocity of an object changes over time. To calculate acceleration in the given scenario, we use the initial and final velocities along with the time taken:\[ a = \frac{v - u}{t} \]where \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is the time. Since the object starts from rest, \( u = 0 \), making the calculation simpler. Let's plug the values into the formula:- Final velocity, \( v = 2.0 \times 10^3 \mathrm{m/s} \)- Initial velocity, \( u = 0 \mathrm{m/s} \)- Time, \( t = 0.96 \mathrm{s} \)This gives us:\[ a = \frac{2.0 \times 10^3 - 0}{0.96} = 2083.33 \mathrm{m/s^2} \]What this means is, the object increases its speed considerably over a short time. Understanding this acceleration completes the puzzle, allowing us to apply Newton's Second Law and further relate everything to the electric field using the known electric force formula.