phys ex. probs

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Physics

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Apr 3, 2024

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Phys Example Problems 1. **Kinematics:** - A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. Calculate the acceleration of the car. - Solution: Using the kinematic equation \(a = \frac{v - u}{t}\), where \(v\) is the final velocity (20 m/s), \(u\) is the initial velocity (0 m/s), and \(t\) is the time (5 seconds), we find \(a = \frac{20 - 0}{5} = 4 \, \text{m/s}^2\). 2. **Newton's Laws of Motion:** - A 10 kg box is pulled along a horizontal surface with a force of 30 N. If the coefficient of friction between the box and the surface is 0.2, calculate the acceleration of the box. - Solution: The net force acting on the box is the difference between the applied force and the force of friction. Using Newton's second law, \(F_{\text{net}} = ma\), where \(F_{\text{net}}\) is the net force, \(m\) is the mass (10 kg), and \(a\) is the acceleration, we find \(30 - \mu mg = ma\), where \(\mu\) is the coefficient of friction and \(g\) is the acceleration due to gravity (9.8 m/s²). Substituting values, we get \(30 - (0.2)(10)(9.8) = 10a\), and solving for \(a\) gives \(a = \ frac{30 - 19.6}{10} = 1.04 \, \text{m/s}^2\). 3. **Work and Energy:** - A 5 kg object is lifted vertically through a height of 10 meters. Calculate the gravitational potential energy gained by the object. - Solution: Gravitational potential energy (\(PE\)) is given by the formula \(PE = mgh\), where \ (m\) is the mass (5 kg), \(g\) is the acceleration due to gravity (9.8 m/s²), and \(h\) is the height (10 m). Substituting values, we get \(PE = (5)(9.8)(10) = 490 \, \text{J}\). 4. **Electricity and Magnetism:** - A charge of 2 C moves through a potential difference of 12 V. Calculate the work done by the electric field on the charge. - Solution: Work done (\(W\)) by the electric field is given by the formula \(W = qV\), where \ (q\) is the charge (2 C) and \(V\) is the potential difference (12 V). Substituting values, we get \ (W = (2)(12) = 24 \, \text{J}\). 5. **Optics:** - A concave mirror has a focal length of 10 cm. Calculate the image distance when an object is placed 20 cm from the mirror. - Solution: Using the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(f\) is the focal length (10 cm), \(d_o\) is the object distance (-20 cm), and \(d_i\) is the image distance, we can solve for \(d_i\). Substituting values, we get \( \frac{1}{10} = \frac{1}{-20} + \frac{1}{d_i} \), and solving for \(d_i\) gives \(d_i = -10 \, \text{cm}\).

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