A student of mass 57.4 kg, starting at rest, slides down a slide 17.2 m long, tilted at an angle of 28.1° with respect to the horizontal. If the coefficient of kinetic friction between the student and the slide is 0.108, find the force of kinetic friction, the acceleration, and the speed she is traveling when she reaches the bottom of the slide. (Enter the magnitudes.)

Explanation:

(a)   Formula to calculate the force of kinetic friction is as follows.

             f =

                =

Putting the given values into the above formula as follows.

          f =

           =

           =

           = 58.54 N

Hence, the force of kinetic friction is 58.54 N.

(b)    Net force experienced by the block will be as follows.

            F =

         ma =

or,         a =                  

                =

                =

                = 4.03

Therefore, the acceleration is 4.03 .

(c)    According to the third equation of motion,

                    =        

                    = 138.63 m/s

Hence, the speed she is traveling when she reaches the bottom of the slide is 138.63 m/s.

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explanation:

f=ma newton

f=100x2.5=250n