Answer:
The frictional force producing this deceleration would have a magnitude of [tex]4\; \rm N[/tex].
Explanation:
The velocity of this object changed by [tex]\Delta v = (-10\; \rm m\cdot s^{-1})[/tex] in [tex]\Delta t = 5\; \rm s[/tex]. The acceleration of this object would be:
[tex]\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:
[tex]\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}[/tex].
([tex]1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}[/tex].)
If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be [tex]4\; {\rm N}[/tex], same as the magnitude of the net force on this object.
