Question 230: Work, Energy and Power & Circular Motion

Question

A force $$\vec{F}=(2+3 x) \hat{i}$$ acts on a particle in the $$x$$ direction where F is in newton and $$x$$ is in meter. The work done by this force during a displacement from $$x=0$$ to $$x=4 \mathrm{~m}$$, is __________ J.

Accepted Answer

To find the work done by a force during a displacement, we can use the formula:
$$W = \int_{x_1}^{x_2} \vec{F} \cdot d\vec{x}$$
Here, the force is given by $$\vec{F} = (2+3x) \hat{i}$$, and we need to find the work done during a displacement from $$x = 0$$ to $$x = 4 \mathrm{~m}$$. Since the force is only in the $$x$$ direction, we can write the integral as:
$$W = \int_{0}^{4} (2+3x) dx$$
Now we can integrate the function with respect to $$x$$:
$$W = \int_{0}^{4} (2+3x) dx = \int_{0}^{4} 2 dx + \int_{0}^{4} 3x dx$$
$$W = \left[ 2x \right]_0^4 + \left[ \frac{3}{2}x^2 \right]_0^4$$
Now we can plug in the limits of integration:
$$W = (2 \cdot 4 - 2 \cdot 0) + \left(\frac{3}{2} \cdot 4^2 - \frac{3}{2} \cdot 0^2 \right)$$
$$W = 8 + 24$$
$$W = 32 \mathrm{~J}$$
So the work done by the force during the displacement from $$x = 0$$ to $$x = 4 \mathrm{~m}$$ is 32 Joules.

Work, Energy and Power & Circular Motion

A force $\vec{F}=(2+3 x) \hat{i} acts on a particle in the x direction where F is in newton and x is in meter. The work done by this force during a displacement from x=0 to x=4 \mathrm{~m}$, is __________ J.

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