A trunk with a weight of 230 N rests on the floor. The coefficient of static fri

Question

A trunk with a weight of 230 N rests on the floor. The coefficient of static friction between the trunk and the floor is 0.25, while the coefficient of kinetic friction is 0.17. What is the minimum magnitude for a horizontal force with which a person must push on the trunk to start it moving?
5.75×101 N



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Once the trunk is moving, what magnitude of horizontal force must the person apply to keep it moving with constant velocity?
3.91×101 N



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If the person continued to push with the force used to start the motion, what would be the acceleration of the trunk?





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Explanation / Answer

force of static friction=N*=230*0.25=57.5

so 57.5 N is the minimum force magnitude in horizontal direction require to make it move (answer 1)

since after it start moving kinetic friction force comes into the action =230*0.17

= 39.1

so 39.1 Nis the force magnitute in horizontal direction to make it move continuously (answer 2)

first force was 57.5 N

now mass=230/9.8=23.469 kg

now

m*a= force - kinetic friction force

23.469*a=57.5-39.1

a = 0.784 m /s^2 (answer 3)

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