The net force acting on a 6.0-kg object is given by \( F_{x}=(10-x) \mathrm{N} \), where \( F_{x} \) is in newtons and \( x \) is in meters. How much work is done on the object as it moves from \( x \) \( =0 \) to \( x=10 \mathrm{~m} \) ? 1) 100 J . 2) 75 J 3) 50 J 4) 25 J 5) 20 J

To find the work done on the object as it moves from \( x = 0 \) to \( x = 10 \) m, we can calculate the work \( W \) done by integrating the net force \( F_{x} \). The force is given by \( F_{x} = (10 - x) \) N. The work done is the integral of the force with respect to displacement: \[ W = \int_{0}^{10} (10 - x) \, dx \] Calculating this integral: \[ W = \left[ 10x - \frac{x^2}{2} \right]_{0}^{10} = \left( 10(10) - \frac{10^2}{2} \right) - \left( 10(0) - \frac{0^2}{2} \right) = (100 - 50) - 0 \] Thus, \[ W = 50 \text{ J}. \] The correct answer is 3) 50 J. Now, here's some extra context if you're curious! When we consider the principles of mechanics in physics, work done by a force is not just a math equation—it has historical roots! The calculation of work comes from the foundational work of scientists like Isaac Newton in the 17th century, who laid the groundwork for classical mechanics. His laws of motion introduced the idea of forces and motion that shape our understanding of the physical world. In real-world applications, understanding work and energy is crucial in fields like engineering and robotics. For instance, when designing a robotic arm, engineers must calculate the work needed to lift objects of various weights. This not only ensures that the motor is powerful enough but also that it can operate efficiently, enhancing the overall performance of the robotic system in practical tasks and innovations!