Final answer:
When a 1.16 kg block is pulled by a 20.54 N force over 2.79 m, opposing friction results in a net force of 18.27 N on the block. This leads to an acceleration of approximately 15.75 m/s². After travelling 2.79 m, the final speed of the block is approximately 9.4 m/s.
Step-by-step explanation:
To calculate the final speed of a 1.16 kg block being pulled across a floor by a 20.54 N force over 2.79 m, we need first to understand the forces at play. The block is acted upon by the force of friction, which opposes the 20.54N force, and also by the force due to acceleration. The force of friction (F.k) is given by the product of the normal force (N) and the coefficient of kinetic friction (μk) - which in this case is 0.2. N is the weight of the block, i.e., its mass times the acceleration due to gravity (9.8 m/s²).
Because the block is moving, we're dealing with kinetic rather than static friction. This means F.k = μk * N = 1.16 kg * 9.8 m/s² * 0.2 ≈ 2.27 N. Now we can calculate the net force on the block, which causes it to accelerate: Net Force = Applied Force - Frictional Force = 20.54 N - 2.27 N = 18.27 N. This net force equals mass times acceleration (F = ma), so we can solve for acceleration: a = F/m = 18.27 N / 1.16 kg ≈ 15.75 m/s².
Finally, to find the speed of the block after it has traveled 2.79 m, we can use the equation of motion v² = u² + 2as, where u is the initial speed (0, as the block starts from rest), a is the acceleration we just found, and s is the distance traveled. Plugging in the values, we get: v² = 0 + 2 * 15.75 m/s² * 2.79 m = 88.39 (m²/s²). So, the final speed v is the square root of this, v = sqrt(88.39) ≈ 9.4 m/s. Therefore, the speed of the block after 2.79 m is 9.4 m/s.
Learn more about Newton's Laws of Motion