- #1

BurpHa

- 44

- 13

- Homework Statement
- A force of 35.0 N is required to start a 6.0-kg box moving

across a horizontal concrete floor. (a) What is the coefficient

of static friction between the box and the floor? (b) If the

35.0-N force continues, the box accelerates at 0.60 m / s ^ 2

What is the coefficient of kinetic friction?

- Relevant Equations
- F_static = \mu_static * mg

I don't understand part (b)

In part (a), I need to calculate the coefficient of the static friction:

mg * \mu_static = 35

58.8 * \mu_static = 35

\mu_static = 35 / 58.8 \approx 0.6

So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot move (since these two forces cancel themselves out.)

However, part (b) says that the box moves at an acceleration of 0.60 m / s ^ 2.

How is it even possible for the box to even start to move and then accelerate?

Thank you for your help.

In part (a), I need to calculate the coefficient of the static friction:

mg * \mu_static = 35

58.8 * \mu_static = 35

\mu_static = 35 / 58.8 \approx 0.6

So from part (a) I know that the force applied is equal to the static friction, meaning that the box cannot move (since these two forces cancel themselves out.)

However, part (b) says that the box moves at an acceleration of 0.60 m / s ^ 2.

How is it even possible for the box to even start to move and then accelerate?

Thank you for your help.