Question

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

released and slides to rest in 5 sec. What is the magnitude of the frictional force
producing this deceleration?

Answer 1

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

Step-by-step 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 &= (\Delta v)/(\Delta t) \\ &= (-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 * (-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.