   Chapter 4, Problem 48P

Chapter
Section
Textbook Problem

A block of mass 12.0 kg is sliding at an initial velocity of 8.00 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.300. (a) What is the force of kinetic friction acting on the block? (b) What is the block’s acceleration? (c) How far will it slide before coming to rest?

(a)

To determine
The force of kinetic friction.

Explanation

Given Info: Mass of the block is 12.0 kg. Initial velocity of the block is 8.00 m/s. Co-efficient of kinetic friction ( μk ) is 0.300.

The normal force balances the weight of the block. Therefore, the formula for normal force is,

N=mg (I)

• m is the mass of the block.
• g is the acceleration due to gravity.

The kinetic force of friction is,

Fk=μkN (II)

Substitute Equation (I) in (II).

Fk=μkmg

Substitute 9.8ms2 for g, 12

(b)

To determine
The acceleration of the block.

(c)

To determine
The sliding distance.

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