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The 6kg box is pushed to the left at a constant speed. The coefficient of friction is 0.78. Solve for the amount of force with which the hand pushes the box.

Sagot :

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Answer:

45.864N

Explanation:

Using the formula

F = nR

n is the coefficient of friction

R is the normal reaction

R = mg

F = nmg

F = 0.78 * 6 * 9.8

F = 45.864N

Hence the amount of force with which the hand pushes the box is 45.864N

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