A 500 g block on a frictionless, horizontal surface is attached to a spring with constant k =8.7 N/m. A second block rests on the first, and the whole system executes simple harmonicmotion with period 1.8 s. When the amplitude of the motion is increased to 35 cm the upperblock just begins to slip. Determine the coefficient of static friction between the blocks
Question
A 500 g block on a frictionless, horizontal surface is attached to a spring with constant k =8.7 N/m. A second block rests on the first, and the whole system executes simple harmonicmotion with period 1.8 s. When the amplitude of the motion is increased to 35 cm the upperblock just begins to slip. Determine the coefficient of static friction between the blocks
Solution
To solve this problem, we need to use the principles of simple harmonic motion and static friction.
Step 1: Find the mass of the block The mass of the block is given as 500 g, but we need it in kilograms for our calculations. So, we convert it to kg by dividing by 1000. m = 500 g = 0.5 kg
Step 2: Find the angular frequency The period of the motion is given as 1.8 s. We can find the angular frequency (ω) using the formula for the period of a simple harmonic motion, T = 2π/ω. Rearranging for ω gives us ω = 2π/T. So, ω = 2π/1.8 = 3.49 rad/s
Step 3: Find the maximum acceleration The maximum acceleration (a_max) in simple harmonic motion occurs at the maximum displacement, which is the amplitude. It can be found using the formula a_max = ω²A, where A is the amplitude. The amplitude is given as 35 cm, which we need to convert to meters by dividing by 100. So, A = 35 cm = 0.35 m. Therefore, a_max = (3.49 rad/s)² * 0.35 m = 4.26 m/s²
Step 4: Find the maximum static friction force The maximum static friction force (f_max) is what keeps the upper block from slipping. It can be found using the formula f_max = m*a_max. So, f_max = 0.5 kg * 4.26 m/s² = 2.13 N
Step 5: Find the coefficient of static friction The coefficient of static friction (μ) can be found using the formula f_max = μN, where N is the normal force. In this case, the normal force is just the weight of the upper block, which is m*g, where g is the acceleration due to gravity (9.8 m/s²). So, N = 0.5 kg * 9.8 m/s² = 4.9 N Rearranging the formula for μ gives us μ = f_max/N. Therefore, μ = 2.13 N / 4.9 N = 0.43
So, the coefficient of static friction between the blocks is 0.43.
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