Force machine pushes cart

Following the push the force machine gives, the cart gains momentum $\Delta p = m\Delta{v} = F\Delta t$ . Using the values in the question statement, we can calculate the initial velocity. $v_i = \frac{F\Delta t}{m} = \frac{1000\times 0.1}{50} = 2\rm ms ^{-1}$ The only unbalanced force acting on the cart between B and C is the frictional force. This force is in the opposite direction of the motion of the cart. Since the acceleration due to this force is parallel to the cart's motion, we can use the kinematic equation $v_f^2 = v_i^2 + 2as$ to solve for the acceleration. We already know that $v_f = 0\rm ms ^{-1}$ and $v_ i = 2\rm ms ^{-1}$ , but we don't know $s$ yet. To calculate $s$ , find the total length of the track between $B = (0,0)$ and $C = (1,\frac{1}{2})$ using the arc length formula. $s = \displaystyle \int_{x = 0}^{1} \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx = \displaystyle \int_{x = 0}^{1} \sqrt{1+x^2}dx$ This integral can be solved using a trigonometric substitution of $x = \tan\theta$ , so $dx = \sec^2\theta d\theta$ . $s = \displaystyle \int_{\theta = \arctan (0)} ^ {\arctan (1)} \sqrt{1+\tan ^2\theta} \sec^2\theta d\theta = \displaystyle \int_{\theta = 0}^{\frac{\pi}{4}} \sec^3\theta d\theta$
This can be solved using integration by parts, yielding $s = \frac{\sqrt{2} + \ln(1+\sqrt{2})}{2}$ Using this, $a = \frac{v_f^2 - v_i^2}{2s} = \frac{0^2 - 2^2}{\sqrt{2} + \ln(1+\sqrt{2})} = \frac{-4}{\sqrt{2} + \ln(1+\sqrt{2})}$ This is negative, which makes sense because the acceleration is in the opposite direction to motion. The friction force is $f = ma = \mu N$ where $N$ is the normal force $N = mg$ . Thus, $\mu = \frac{ma}{N} = \frac{ma}{mg} = \frac{a}{g} = \boxed{\frac{4}{g(\sqrt{2} + \ln(1+\sqrt{2}))}}$ The acceleration is taken as positive because the coefficient of friction must be positive. $a + b = 1 + 2 = \boxed{3}$