Question

A force F=1+2x acts on a body in x-direction where x is in metres and F in newton. Find the work done in displacing the body from x=0 to x=6 m.

• A. 14 J
• B. 18 J
• C. 16 J
• D. None of these

Answer 1

Answer: (D)

42 J

Answer 2

The work done by a variable force $F(x)$ acting on a body in displacing it from $x_1$ to $x_2$ is given by the integral:

$W = \int_{x_1}^{x_2} F(x) dx$

In this problem, the force is given by $F(x) = 1 + 2x$ newton. The displacement is from $x_1 = 0$ m to $x_2 = 6$ m.

Substituting the given force and limits into the formula, we get:

$W = \int_{0}^{6} (1 + 2x) dx$

Now, we evaluate the definite integral:

$W = \left[ x + \frac{2x^2}{2} \right]_{0}^{6}$ $W = \left[ x + x^2 \right]_{0}^{6}$

Evaluate the expression at the upper limit ($x=6$) and subtract the evaluation at the lower limit ($x=0$):

$W = (6 + 6^2) - (0 + 0^2)$ $W = (6 + 36) - (0)$ $W = 42 - 0$ $W = 42 \text{ J}$

The calculated work done is 42 J. Therefore, the correct option is (4) None of these.