Problemas
A block of mass 2.0 kg is initially at rest on a horizontal surface. The block is pulled a distance of 10 m in 3.0 s by an applied force. The coefficients of static and kinetic friction between the surface and the object are mu _(s)=0.45 and mu _(k)=0.30 respectively. The magnitude of the applied force is most nearly A square B square C square square D
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To solve this problem, we need to calculate the magnitude of the applied force required to pull the block a distance of 10 m in 3.0 s, taking into account the forces of friction.Given information:- Mass of the block: 2.0 kg- Distance pulled: 10 m- Time taken: 3.0 s- Coefficient of static friction: μs = 0.45- Coefficient of kinetic friction: μk = 0.30Step 1: Calculate the acceleration of the block.Since the block is initially at rest, the acceleration can be calculated using the kinematic equation:v^2 = u^2 + 2aswhere v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.Since the block is initially at rest, u = 0.v^2 = 2asa = v^2 / (2s)a = (10 m / 3.0 s)^2 / (2 × 10 m)a = 1.85 m/s^2Step 2: Calculate the force of friction.The force of friction is given by:F_friction = μ × Nwhere μ is the coefficient of friction and N is the normal force.Since the block is on a horizontal surface, the normal force is equal to the weight of the block.N = m × gwhere m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).N = 2.0 kg × 9.8 m/s^2N = 19.6 NF_friction = μ × NF_friction = 0.30 × 19.6 NF_friction = 5.88 NStep 3: Calculate the applied force.The applied force must overcome the force of friction and provide the necessary acceleration to the block.F_applied = F_friction + m × aF_applied = 5.88 N + 2.0 kg × 1.85 m/s^2F_applied = 5.88 N + 3.7 NF_applied = 9.58 NTherefore, the magnitude of the applied force is most nearly 9.0 N (Option B).