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))

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge