Question

A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly

Answer 1

Answer:

The frictional force producing this deceleration would have a magnitude of $4\; \rm N$.

Explanation:

The velocity of this object changed by $\Delta v = (-10\; \rm m\cdot s^{-1})$ in $\Delta t = 5\; \rm s$. The acceleration of this object would be:

$\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}$.

Let $m$ denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:

$\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}$.

($1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}$.)

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 $4\; {\rm N}$, same as the magnitude of the net force on this object.