a block of mass 5kg initially at rest at the origin is acted on by a force along the position X direction represented by F=(20+5x)N calculate the work done by the force during the displacement of the block from x=0 at x=4m
To calculate the work done by the force during the displacement of the block, we can use the formula:
Work = Integral(F * dx)
Given that the force is represented by F = (20 + 5x) N, and the displacement of the block is from x = 0 to x = 4 m, we can substitute these values into the formula:
Work = Integral((20 + 5x) * dx)
Now, we need to integrate this expression with respect to x over the displacement range (0 to 4):
Work = ∫(20 + 5x) dx
Integrating with respect to x, we get:
Work = 20x + (5/2)x^2 + C
where C is the constant of integration.
To find the value of C, we need to evaluate the expression at the initial position x = 0:
Work at x = 0 = 20(0) + (5/2)(0)^2 + C
Work at x = 0 = C
Since the block is initially at rest, the work done at x = 0 is zero. Therefore, C = 0.
Substituting the value of C back into the equation, we get:
Work = 20x + (5/2)x^2
Now, we can calculate the work done during the displacement from x = 0 to x = 4:
Work = 20(4) + (5/2)(4)^2
Work = 80 + (5/2)(16)
Work = 80 + 40
Work = 120 J
Therefore, the work done by the force during the displacement of the block from x = 0 to x = 4 m is 120 Joules.
integral F dot dx from x = 0 to x = 4
int [ 20 + 5 x ] dx = 20 int dx + 5 int x dx
= 20 (4 - 0) + 5 (.5 * 4^2 - .5 * 0^2)
= 80 + 40
= 120 Joules