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# Lecture 13 ## Work Done by a Constant Force - Consider a particle being displaced from A to B along a straight line by a constant force $\vec{F}$. $\qquad \vec{d} = \vec{r}_B - \vec{r}_A$ The work done by the force $\vec{F}$ is defined to be $$ W = \vec{F} \cdot \vec{d} = Fd...

# Lecture 13 ## Work Done by a Constant Force - Consider a particle being displaced from A to B along a straight line by a constant force $\vec{F}$. $\qquad \vec{d} = \vec{r}_B - \vec{r}_A$ The work done by the force $\vec{F}$ is defined to be $$ W = \vec{F} \cdot \vec{d} = Fd \cos{\theta} $$ where $\theta$ is the angle between $\vec{F}$ and $\vec{d}$. - Work is a scalar quantity. The SI unit of work is the joule (J). $$ 1 \text{ joule } = 1 \text{ J } = 1 \text{ N.m} $$ - The work done is positive if $0 \leq \theta < 90^\circ$. - The work done is negative if $90^\circ < \theta \leq 180^\circ$. - The work done is zero if $\theta = 90^\circ$. ## Example: A particle moving in the xy plane undergoes a displacement given by $\vec{d} = (2.0\hat{i} + 3.0\hat{j}) \text{ m}$ as a constant force $\vec{F} = (5.0\hat{i} + 2.0\hat{j}) \text{ N}$ acts on the particle. a) Calculate the magnitude of the force and the displacement. b) Calculate the work done by the force. $$ \vec{d} = (2.0\hat{i} + 3.0\hat{j}) \text{ m}, \qquad \vec{F} = (5.0\hat{i} + 2.0\hat{j}) \text{ N} $$ a) $$ \begin{aligned} |\vec{F}| &= \sqrt{5^2 + 2^2} = 5.4 \text{ N} \\ |\vec{d}| &= \sqrt{2^2 + 3^2} = 3.6 \text{ m} \end{aligned} $$ b) $$ \begin{aligned} W &= \vec{F} \cdot \vec{d} = (5.0\hat{i} + 2.0\hat{j}) \cdot (2.0\hat{i} + 3.0\hat{j}) \\ &= (5.0)(2.0) + (2.0)(3.0) = 16 \text{ J} \end{aligned} $$ ## Work Done by a Varying Force - If the force varies during the displacement, then the work done by the force is $$ W = \int_{x_i}^{x_f} F_x dx $$ where $F_x$ is the component of the force in the direction of the displacement. - In three dimensions, $$ W = \int_{r_i}^{r_f} \vec{F}(\vec{r}) \cdot d\vec{r} = \int_{x_i}^{x_f} F_x dx + \int_{y_i}^{y_f} F_y dy + \int_{z_i}^{z_f} F_z dz $$ ### Example A force acting on a particle varies with $x$ as shown in Figure. Calculate the work done by the force on the particle as it moves from $x = 0$ to $x = 6.0 \text{ m}$. **Solution** The work done by the force is equal to the area under the curve. $W = \text{Area of triangle } (0 \rightarrow 4) + \text{Area of rectangle } (4 \rightarrow 6)$ $W = \frac{1}{2} (4)(5) + (2)(5) = 10 + 10 = 20 \text{ J}$ ## Work Done by a Spring - The force exerted by a spring is given by Hooke's Law: $$ F_x = -kx $$ where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. - The work done by the spring force is $$ W_s = \int_{x_i}^{x_f} F_x dx = \int_{x_i}^{x_f} -kx dx = -\frac{1}{2} kx^2 \Big|_{x_i}^{x_f} = \frac{1}{2} kx_i^2 - \frac{1}{2} kx_f^2 $$ - The work done by an applied force is $$ W_a = \int_{x_i}^{x_f} F_x dx = \int_{x_i}^{x_f} kx dx = \frac{1}{2} kx^2 \Big|_{x_i}^{x_f} = \frac{1}{2} kx_f^2 - \frac{1}{2} kx_i^2 $$