A force of 57.1 Newtons is applied for 0.977 meters; what work was done on the object? (1 point) \[ 55.8 \mathrm{~J} \] 556.1 J 58.1 J 58.4 J

To find the work done on the object, we can use the formula for work: \[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \] Here, the force is 57.1 Newtons, the distance is 0.977 meters, and since the force is applied in the direction of the movement, \(\cos(0°) = 1\). Thus, \[ \text{Work} = 57.1 \, \text{N} \times 0.977 \, \text{m} \times 1 = 55.8 \, \text{J} \] So, the correct answer is **55.8 J**. In practical life, understanding work is crucial! For example, when you're pushing a heavy object across the floor, the work you do depends not only on how hard you push (the force) but also on how far you move it. The energy transfer is what helps you move things, and it's not just a physics concept; it's about getting things done in daily activities! Additionally, when solving work problems, a common mistake is forgetting to account for the angle between force and displacement. Always remember that if the force is at an angle, you need to use the cosine of that angle. In this case, since the force and movement were aligned, it made it simpler—no mixing up those cosines!