# physics help w/ question…?

a box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Find ,uk between the box and the floor…?

• odu83
1 month ago

The normal force is

325+425*sin(35.2)

therefore

(325+425*sin(35.2))*µk=425*cos(35.2)

µk=0.61

j

• ukmudgal
1 month ago

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The weight of box of books = w = 325 N

The box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal.

Force that pushes the box = F =425 N

Angle which the force makes with horizontal =O =35.2 degree

Horizontal component of force = FcosO=425*0.8171=347.29

Horizontal component of force = 347.29 N

Vertical downward component of force = FsinO=425*0.5764=244.98

Vertical downward component of force = 244.98 N

The box moves at a constant velocity across the floor .

Resultant force on the box is zero

Normal reaction = R= Weight + vertical downward component of force

Normal reaction = R= 325+244.98=569.98 N

The force of friction balances the horizontal component of force because the box moves at a constant velocity across the floor .

The force of friction =mu*Normal reaction=mu*569.98

The force of friction = the horizontal component of force

mu*569.98 = 347.29

mu =347.29 /569.98=0.6092

The coefficient of kinetic friction between box and floor is 0.6092

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