Answer:
the force must be larger than F= 490N for example for a speed of 1m/s for 10s the force is 590N
Explanation:
The motion of the truck must be bigger than the Fk of friction that is opposite of the force both person are doing so:
[tex]F-F_{k}=m*a[/tex]
Knowing the truck must be accelerated to get a constant speed the acceleration can be determinate by
[tex]V_{f}=V_{o}+a*t[/tex]
It is at rest so
[tex]V_{f}=a*t[/tex]
[tex]a=\frac{V_{f}}{t}[/tex]
The force net of both person is
[tex]F<F_{k}[/tex]
[tex]F=u_{k}*m*g[/tex]
[tex]F=0.5*100kg*9.8\frac{m}{s^2}[/tex]
[tex]F=490N[/tex]
Asume for example that the speed they want is about 1 m/s for 10 s so the force have to be larger than 490N
[tex]F=m*(\frac{V_{f}}{t}+m*g )[/tex]
[tex]F=100kg*(1\frac{m}{s}}+0.5*9.8\frac{m}{s^2})[/tex]
[tex]F=590N[/tex]
