Answer:
[tex]500\; {\rm J}[/tex].
Explanation:
If the force on the object is constant, work done on the object would be equal to the product of force and the displacement in the direction of the motion:
[tex]W = F\, x[/tex].
However, since the force in this question varies with position, finding the total work done on the object requires integrating force with respect to displacement:
[tex]\begin{aligned} W &= \int\limits_{0}^{10} F(x)\, d x \\ &= \int\limits_{0}^{10} (10\, x)\, d x \\ &= \left[5\, x^{2}\right]_{0}^{10} \\ &= 5\, \left(10^{2}\right) - 5\, \left(0^{2}\right) \\ &= 500\end{aligned}[/tex].
In other words, the work done on this object would be [tex]500\; {\rm J}[/tex] ([tex]1\; {\rm J} = 1\; {\rm N\cdot m}[/tex]) when it moves from [tex]x = 0\; {\rm m}[/tex] to [tex]x = 10\; {\rm m}[/tex] in the direction of this variable force.
The work done on the 6.0-kg object as it moves from x = 0 to x = 10 m under a variable force Fx = (10 x) N is calculated through integration, resulting in 500 joules.
The question asks to calculate the work done on a 6.0-kg object as it moves from x = 0 to x = 10 m under the influence of a variable force Fx = (10 x) N. In physics, work is defined by the equation W = F · d, where W is work, F is the constant force, and d is the displacement.
However, because the force in this scenario depends on the position (x), we cannot use the simple equation for work directly since the force isn't constant. Instead, we need to integrate the force function over the displacement:
Work (W) performed is obtained through the integral of F(x) with respect to x from 0 to 10 m:
[tex]W = ∫010 (10x) dx = ½(10)(10^2) - ½(10)(0^2) = 500 J[/tex]
The integral of 10x from 0 to 10 m is the area under the force versus position graph, which corresponds to a triangle with a base of 10 m and a height of 100 N. Therefore, the work done is half the product of the base and the height, which is 500 joules (J).
