a 10 kg box is sliding to the right on a horizontal surface .the coefficient of

Question

a 10 kg box is sliding to the right on a horizontal surface .the coefficient of kinetic friction between the box and the surface is .6. A force F1= N (up and to the right at an angle of 25 degrees above the horizontal) is applied to the box as well as force F2. What is x component of F1? Y component of F1? Magnitude of box's weight? Magnitude of normal force on box? Magnitude of friction force on box? What is F2? a 10 kg box is sliding to the right on a horizontal surface .the coefficient of kinetic friction between the box and the surface is .6. A force F1= N (up and to the right at an angle of 25 degrees above the horizontal) is applied to the box as well as force F2. What is x component of F1? Y component of F1? Magnitude of box's weight? Magnitude of normal force on box? Magnitude of friction force on box? What is F2? a 10 kg box is sliding to the right on a horizontal surface .the coefficient of kinetic friction between the box and the surface is .6. A force F1= N (up and to the right at an angle of 25 degrees above the horizontal) is applied to the box as well as force F2. What is x component of F1? Y component of F1? Magnitude of box's weight? Magnitude of normal force on box? Magnitude of friction force on box? What is F2?

Explanation / Answer

the x component of F1 is = F1cos(theta) = F1 cos25 [but unfortunately the F1 VALUE MISSING]

and the y component of F1 = F1 sin25

the magnitude of the box weight = mg = 10x9.8 = 98 Newtons

magnitude of normal force on box = mg - F1 sin25 = 98 - F1sin25 newtons

magnitude of friction force = 0.6 mg = 0.6x98 = 58.8 newtons

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