   Chapter 4, Problem 77AP

Chapter
Section
Textbook Problem

A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 7.00 m/s. If the coefficient of friction between the sled’s runners and the snow is 0.050 0 and the boy and sled together weigh 600. N, how far does the sled travel on the level surface before coming to rest? To determine
The distance travelled by the sled before coming to rest.

Explanation

Given Info: Total weight is 600 N. Speed of the sled is 7.00 m/s. The co-efficient of friction is 0.050.

The formula for the friction force is,

Fk=μkN (I)

• μk is the co-efficient of friction.
• N is the normal force

The normal force is equal to the total weight. The force of friction is equal to the horizontal force. Therefore,

ma=Fk

• m is the total mass.
• a is the acceleration.

Substitute W/g for m and re-arrange to get a.

a=Fk(Wg)=(FkgW)

Substitute Equation (II) in the above equation get a.

a=(μkNgW)

Substitute W for N in the above equation.

a=μkg (II)

From Newton’s equation of motion, the formula for distance is,

s=v2u22a (III)

• v is the final velocity

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