Solved Problem on Dynamics
Português     English

A body of mass 3 kg, initially at rest, is on a frictionless horizontal surface. A horizontal force of constant intensity equal to 4.5 N acts on the body for 20 s, find:
a) What is the acceleration of the body while the force acts?
b) What is the speed of the body when the force stops acting?
c) What is the distance traveled by the body until the force stops acting?

Problem data
• mass of the body:    m = 3 kg;
• initial speed of the body:    v0} = 0;
• force applied to the body:    F = 4.5 N;
• interval of time in which the force acts on the body:       t = 20 s.
Problem diagram

A reference frame oriented to the right is adopted, with the body initially at rest at the origin (figure 1).

figure 1

Solution

a) Drawing a diagram of the body with the forces acting on it (figure 2)

Horizontal direction
• $$\vec{F}$$ : force applied to the body.
Vertical direction
• $$\vec{N}$$ : normal force of the surface on the body;
• $$\vec{P}$$ : weight.

figure 2
In the vertical direction, there is no motion, normal force and weight cancel out.
In the horizontal direction, the only force that acts is the force applied to the body, the resultant in this direction will be the force itself, $$\vec{F}$$, applying Newton's Second Law we have
$\bbox[#99CCFF,10px] {\vec{F}=m\vec{a}}$
$4.5=3\;a\\ a=\frac{4,5}{3}$
$\bbox[#FFCCCC,10px] {a=1.5\;\text{m/s}^{2}}$

b) The body is under constant acceleration, applying the equation of velocity as a function of time
$\bbox[#99CCFF,10px] {v=v_{0}+at}$
$v=0+1.5\times20$
$\bbox[#FFCCCC,10px] {v=30\;\text{m/s}}$

c) Applying the equation of the displacement as a function of time we have
$\bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}\;t^{2}}$
$S=0+0\times20+\frac{1.5}{2}\times20^{2}\\ S=\frac{1.5}{2}\times400\\ S=1.5\times200$
$\bbox[#FFCCCC,10px] {S=300\;\text{m}}$