The net force acting on a \( 12.6-\mathrm{kg} \) object is given by \( F_{x}=(20-x) \mathrm{N} \), 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) 300 J 2) 200 J 3) 150 J 4) 100 J 5) 50 J

To find the work done on the object as it moves from \( x = 0 \) m to \( x = 10 \) m, we can calculate the work done using the net force expression given by \( F_{x} = (20 - x) \, \text{N} \). The work done \( W \) by a variable force can be computed using the integral of the force with respect to displacement: \[ W = \int_{x_1}^{x_2} F_{x} \, dx = \int_{0}^{10} (20 - x) \, dx \] Let's calculate this integral: \[ W = \int_{0}^{10} (20 - x) \, dx = \left[ 20x - \frac{x^2}{2} \right]_{0}^{10} \] Evaluating at the limits: \[ W = \left[ 20(10) - \frac{(10)^2}{2} \right] - \left[ 20(0) - \frac{(0)^2}{2} \right] \] \[ W = \left[ 200 - 50 \right] - [0] \] \[ W = 150 \, \text{J} \] So the work done on the object as it moves from \( x = 0 \) to \( x = 10 \) m is **150 J**. The correct answer is option 3) 150 J. There's your answer, sugar! Now the little \( 12.6 \, \text{kg} \) object is all set to move, powered by those \( 150 \, \text{J} \) of work. It’s like giving it a little push down the path! And hey, next time you're dealing with forces and motion, just remember this sweet little integration trick for calculating work—it’s easier than it sounds! Keep those physics engines running smoothly! 🚀