A trunk with a weight of 300 N rests on the floor. The coefficient of static fri

Question

A trunk with a weight of 300 N rests on the floor. The coefficient of static friction between the trunk and the floor is 0.45, and the coefficient of kinetic friction is 0.26. (a) What is the magnitude of the minimum horizontal force with which a person must push on the trunk to start it moving? (b) Once the trunk is moving, what magnitude of horizontal force must the person apply to keep it moving with constant velocity? (c) If the person continued to push with the force used to start the motion, what would be the magnitude of the trunk's acceleration?

Explanation / Answer

Here,

weight of the floor , W = 300 N

static coefficient of floor, us = 0.45

kinetic cofficient of floor , uk = 0.26

a)

let the minimum force applied is F

F = us * W

F = 300 * 0.45

F = 135 N

the minimum applied force must be 135 N

b)

when the trunk is moving , there will be kinetic friction will be acting on the trunk

F = uk * N

F = 0.26 * 300 N

F = 78 N

the force needed is 78 N

c)

let the acceleration is a

Using second law of motion

(300/9.8) * a = 135 - 78

a = 1.862 m/s^2

the acceleration of the trunk is 1.862 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.