Question

A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 380 N . For the first 14 m the floor is frictionless, and for the next 14 m the coefficient of friction is 0.35.

What is the final speed of the crate?

Answer 1

here,

for the first 14 m

the accelration ,a = net force /effective mass

a = ( F )/m

a = ( 380 )/110

a = 3.45 m/s^2

final speed , v0 = sqrt(2 * a * s)

v0 = sqrt(2 * 3.45 * 14) = 9.83 m/s

for the next 14 m

the accelration ,a = net force /effective mass

a = ( F - uk * m * g)/m

a = ( 380 - 0.35 * 9.81 * 110)/110

a = 0.021 m/s^2

let the final speed be v

v^2 - v0^2 =2 * a * s

v^2 - 9.83^2 = 2 * 0.021 * 14

solving for v

v = 9.86 m/s

the final speed of crate is 9.86 m/s