Question

A trunk of mass 16 kg is on the floor. The trunk has a very small initial speed. The acceleration of gravity is 9.8 m/s 2 . What constant horizontal force pushing thetrunk is required to give it a velocity of 10 m/s in 20 s if the coefficient of sliding friction between the trunk and the floor is 0.57? Answer in units of N.

Answer 1

Answer:

F=96 N

Explanation:

First of all, we need to calculate the acceleration, that's given by:

$a=\frac{V_f-V_i}{t}$

the problem states that the initial velocity is very small, so we can assume it as zero, so:

$a=\frac{10m/s}{20s}\\\\a=0.50m/s^2$

according to Newton's second law:

$\sum F=m.a\\-F_f+F=m.a\\F_f=\µ.m.g\\F_f=0.57*16kg*9.8m/s^2\\F_f=89N\\\\so:\\\\F=13kg*0.50m/s^2+89N\\F=96N$