The drawing shows a 25.0-kg crate that is initially at rest.Note that the view i

Question

The drawing shows a 25.0-kg crate that is initially at rest.Note that the view is one looking down on the top of the crate. Twoforces, F1 and F2 are applied to the crate, and it begins to move.The coefficient of kinetic friction between the crate and the flooris k=0.350. Determine the magnitude and direction (relativeto the x axis) of the acceleration of the crate. The drawing shows a 25.0-kg crate that is initially at rest.Note that the view is one looking down on the top of the crate. Twoforces, F1 and F2 are applied to the crate, and it begins to move.The coefficient of kinetic friction between the crate and the flooris k=0.350. Determine the magnitude and direction (relativeto the x axis) of the acceleration of the crate.

Explanation / Answer

   to get the result consider the free bodydiagram as shown
   the sum of the two applied forces will be    F = F1 + F2    the x component of the sum will be    Fx = F1cos55.0o + F2        = (88.0 N)cos55.0o + (54.0 N)        = ........ N    the y component of the sum willbe    Fy = F1sin55.0o        = (88.0 N)sin55.0o        = ........N    so the magnitude of the force willbe    F = [(Fx)2+ (Fy)2]       = ......... N    in the given question we can see that as thecrate starts from rest it moves along the direction    of F    the kinetic frictional forcefk opposes the motion, so it points oppositeto F.    the net force acting on the crate is the sumof F and fk.    the magnitude a of thecrate’s acceleration is equal to the magnitude SF of the netforce divided    by the mass m of the crate    so then we get    a = F / m       = (- fk + F) /m    the kinetic frictional force is givenby    fk = kFN    FN = m g    so the acceleration becomes    a = (- k m g + F) / m       = .......... m /s2    the angle will be     = tan-1(Fy /Fx)       = ........o        = (88.0 N)sin55.0o        = ........N    so the magnitude of the force willbe    F = [(Fx)2+ (Fy)2]       = ......... N    in the given question we can see that as thecrate starts from rest it moves along the direction    of F    the kinetic frictional forcefk opposes the motion, so it points oppositeto F.    the net force acting on the crate is the sumof F and fk.    the magnitude a of thecrate’s acceleration is equal to the magnitude SF of the netforce divided    by the mass m of the crate    so then we get    a = F / m       = (- fk + F) /m    the kinetic frictional force is givenby    fk = kFN    FN = m g    so the acceleration becomes    a = (- k m g + F) / m       = .......... m /s2    the angle will be     = tan-1(Fy /Fx)       = ........o        = ........N    so the magnitude of the force willbe    F = [(Fx)2+ (Fy)2]       = ......... N    in the given question we can see that as thecrate starts from rest it moves along the direction    of F    the kinetic frictional forcefk opposes the motion, so it points oppositeto F.    the net force acting on the crate is the sumof F and fk.    the magnitude a of thecrate’s acceleration is equal to the magnitude SF of the netforce divided    by the mass m of the crate    so then we get    a = F / m       = (- fk + F) /m    the kinetic frictional force is givenby    fk = kFN    FN = m g    so the acceleration becomes    a = (- k m g + F) / m       = .......... m /s2    the angle will be     = tan-1(Fy /Fx)       = ........o    the angle will be     = tan-1(Fy /Fx)       = ........o

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