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Chapter 4: Problem 422

A ball of mass \(5 \mathrm{~kg}\) is striding on a plane with initial velocity of \(10 \mathrm{~m} / \mathrm{s}\). If co-efficient of friction between surface and ball is \((1 / 2)\), then before stopping it will describe \(\ldots \ldots\) \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(12.5 \mathrm{~m}\) (B) \(5 \mathrm{~m}\) (C) \(7.5 \mathrm{~m}\) (D) \(10 \mathrm{~m}\)

Short Answer

Expert verified
The ball will describe a distance of \(10 \mathrm{~m}\) before stopping.

Step by step solution

01

Find the Friction Force

First, we need to calculate the friction force acting on the ball. This force can be determined using the following formula: \[ F_{friction} = \mu \cdot F_{normal} \] Where \(F_{friction}\) is the friction force, \(\mu\) is the coefficient of friction, and \(F_{normal}\) is the normal force. Since the ball is moving on a horizontal surface, the normal force equals the gravitational force acting on it: \[ F_{normal} = m \cdot g \] Now, we can plug in the given values and find the friction force: \[ F_{friction} = \frac{1}{2} \cdot 5 \mathrm{~kg} \cdot 10 \mathrm{~m/s^2} = 25 \mathrm{~N} \]
02

Calculate the Deceleration

Next, we will find the deceleration caused by the friction force acting on the ball. To do this, we'll use Newton's second law of motion: \[ F = m \cdot a \] Where \(F\) is the net force acting on the object, \(m\) is the mass of the object, and \(a\) is the acceleration (or in our case, deceleration). In this case, the friction force is the only force acting on the ball, and it acts opposite to its direction of motion. So, we can write: \[ a = \frac{F_{friction}}{m} \] Plugging in our values, we get: \[ a = \frac{25 \mathrm{~N}}{5 \mathrm{~kg}} = 5 \mathrm{~m/s^2} \]
03

Find the Time to Stop

Now, we will find the time it takes for the ball to stop due to the deceleration caused by friction. Using the formula for uniformly accelerated motion, we have: \[ v_f = v_i - a \cdot t \] Where \(v_f\) is the final velocity (0 m/s in this case, as the ball stops), \(v_i\) is the initial velocity (10 m/s), \(a\) is the deceleration (5 m/s^2), and \(t\) is the time. Solving for \(t\), we get: \[ t = \frac{v_i - v_f}{a} = \frac{10 \mathrm{~m/s} - 0 \mathrm{~m/s}}{5 \mathrm{~m/s^2}} = 2 \mathrm{~s} \]
04

Calculate the Distance Traveled

Finally, we will find the distance the ball travels before stopping using the formula for uniformly accelerated motion: \[ d = v_i \cdot t - \frac{1}{2} \cdot a \cdot t^2 \] Plugging in the values we have found, we get: \[ d = 10 \mathrm{~m/s} \cdot 2 \mathrm{~s} - \frac{1}{2} \cdot 5 \mathrm{~m/s^2} \cdot (2 \mathrm{~s})^2 \] \[ d = 20 \mathrm{~m} - 10 \mathrm{~m} = 10 \mathrm{~m} \] The ball will describe a distance of \(10 \mathrm{~m}\) before stopping. Hence, the correct answer is (D).

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