A5.86 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 14.4 n. the coefficient of kinetic friction is 0.14. the acceleration of gravity is 9.8 m/s 2 . find the speed of the block after it has moved 4.98 m. answer in units of m/s.

3.3m/s

Explanation:

1. Find the acceleration a:

Given: normal force: N, the horizontal force: F, friction coefficient: μ, mass: m, acceleration of gravity: g. The sum of all forces in the horizontal plane is:

∑F = ma = F - μN = F - μmg

Solve for the acceleration a:

(i) a = F/m - μg

2. Use the equations of motion to calculate the velocity v for a uniform acceleration a, time t and initial velocity v₀:

(ii) v = at + v₀

position x, initial position x₀:

(iii) x = 0.5at² + v₀t + x₀

Eliminate time t by plugging in eq. (ii) into eq. (iii) with v₀ = 0 and x₀ = 0:

(iv) x = 0.5v²/a

Solve for v:

(v) v = √(2xa) = √(2*x*( F/m - μg)) = √(2*4.89 * (14.4/5.86 - 0.14*9.8))

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.