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A5.00kgstone is rubbed across the horizontal ceiling of a cave passageway (Fig. 6-48). If the coefficient of kinetic friction is0.65and the force applied to the stone is angled atθ=70.0°, what must the magnitude of the force be for the stone to move at constant velocity?

Short Answer

Expert verified

F=118 N

Step by step solution

01

Given data

  • Mass of stone,m=5.00 kg.
  • Coefficient of friction,μk=0.65.
  • The angle of force,θ=70.0o
02

To understand the concept

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.

Draw the free body diagram of the stone. Then, applying Newton’s second law, the given problem can be solved.

03

Draw the free body diagram and write force equations

The free-body diagram for the stone is shown below, with Fbeing the force applied to the stone, FNthe normal downward force of the ceiling on the stone, mgthe force of gravity, and fthe force of friction. We take the +xdirection to be horizontal to the right and the +ydirection to be up.

The equations for thexand theycomponents of the force according to Newton's second law are:

role="math" localid="1661240446674" Fx=Fcosθf=ma

Fy=FsinθFNmg=0

04

Calculate the magnitude of the force for the stone to move at constant velocity 

Friction can be written as,

f=μkFN

f=μk(Fsinθmg)

This expression is substituted forfin the first equation to obtain as:

Fcosθμk(Fsinθmg)=ma.

For a=0, the force is,

F=μkmgcosθμksinθ

With μk=0.65,m=5.0 kg, and θ=70o, we can write the above expression as,

F=0.65×5.0 kg×9.8 m/s2cos70o0.65sin70oF=118 N

Thus, the magnitude of the force must be 118 N for the stone to move at constant velocity.

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