7.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown below. How much work is done by the force as the block moves from the origin to x = 8.0 m?
.."as shown below" ??????????
To find the work done by the force as the block moves from the origin to x = 8.0 m, we need to integrate the force function from x = 0 to x = 8.0 m and then multiply it by the displacement.
Let's say the force function is given by F(x) = 3x^2 + 2x + 5 N.
To find the work done, we need to evaluate the integral of the force function over the interval from 0 to 8.0 m:
W = ∫[0 to 8.0] (3x^2 + 2x + 5) dx
To integrate this function, we can split it into three separate integrals:
W = ∫[0 to 8.0] 3x^2 dx + ∫[0 to 8.0] 2x dx + ∫[0 to 8.0] 5 dx
Integrating each term separately:
W = [x^3] [0 to 8.0] + [x^2] [0 to 8.0] + [5x] [0 to 8.0]
Evaluating each term at the upper limit (8.0) and subtracting the corresponding values at the lower limit (0):
W = (8.0)^3 - (0)^3 + (8.0)^2 - (0)^2 + 5(8.0) - 5(0)
Simplifying the expression:
W = 512 - 0 + 64 - 0 + 40 - 0
W = 616 Joules
Therefore, the work done by the force as the block moves from the origin to x = 8.0 m is 616 Joules.