To determine the work done by friction on the bead, we need to follow a systematic approach. Let's break down the problem step-by-step. ### Step 1: Understand the Problem We have a bead with a mass of \( m = 0.01 \, \text{kg} \) that comes to rest due to friction. The coefficient of friction \( \mu = 0.1 \) and the bead travels a distance of \( d = 0.2 \, \text{m} \). We need to find the work done by the friction force. ### Step 2: Calculate the Normal Force The normal force \( N \) acting on the bead is equal to its weight when it is on a horizontal surface. The weight \( W \) can be calculated using the formula: \[ W = m \cdot g \] where \( g = 10 \, \text{m/s}^2 \) (the acceleration due to gravity). Substituting the values: \[ W = 0.01 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.1 \, \text{N} \] Thus, the normal force \( N = 0.1 \, \text{N} \). ### Step 3: Calculate the Frictional Force The frictional force \( F_f \) can be calculated using the formula: \[ F_f = \mu \cdot N \] Substituting the values: \[ F_f = 0.1 \cdot 0.1 \, \text{N} = 0.01 \, \text{N} \] ### Step 4: Calculate the Work Done by Friction The work done by the friction force \( W_f \) can be calculated using the formula: \[ W_f = F_f \cdot d \] Since friction acts in the opposite direction to the motion, the work done by friction will be negative: \[ W_f = -F_f \cdot d \] Substituting the values: \[ W_f = -0.01 \, \text{N} \cdot 0.2 \, \text{m} = -0.002 \, \text{J} \] Thus, the work done by friction is: \[ W_f = -2 \times 10^{-3} \, \text{J} \] ### Step 5: Determine the Correct Option The magnitude of the work done by friction is \( 2 \times 10^{-3} \, \text{J} \). Since the question asks for the work done by friction (which is negative), we can refer to the absolute value when matching with the options provided. The correct option is: **B. \( 2 \times 10^{-3} \)**. ### Step 6: Analyze Other Options - **Option A: \( 2 \times 10^{-4} \)**: This value is too small and does not correspond to the calculated work done by friction. - **Option C: \( 2 \times 10^{-1} \)**: This value is too large and does not match our calculations. - **Option D: \( 2 \times 10^{2} \)**: This is an extremely large value and is not relevant to the problem. ### Summary - The work done by friction is calculated using the frictional force and the distance traveled. - The frictional force is determined by the coefficient of friction and the normal force. - The correct answer is \( 2 \times 10^{-3} \, \text{J} \), which corresponds to option B. - Always remember that work done against friction is negative, as it opposes the motion. ### Revision Summary - Work done by friction is calculated as \( W_f = -F_f \cdot d \). - The frictional force is \( F_f = \mu \cdot N \). - Ensure to use the correct signs when calculating work done against friction. - The correct answer for this problem is \( 2 \times 10^{-3} \, \text{J} \) (Option B).