My NotesO Ask Your Teacher I. :.3 points SerCP11 4P022. A student of mass 55.4 k

Question

My NotesO Ask Your Teacher I. :.3 points SerCP11 4P022. A student of mass 55.4 kg, starting at rest, slides down a slide 19.2 m long, tilted at an angle of 22.1° with respect to the horizontal. If the coefficient of kinetic friction between the student and the slide is 0.123, 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.) HINT (a) the force of kinetic friction (in N) (b) the acceleration (in m/s2) m/s2 (c) the speed she is traveling (in m/s) m/s Need Help?Watch

Explanation / Answer

(a) Force of kinetic friction, Fk = u*mg*cos22.1 = 0.123*55.4*9.81*cos22.1 = 61.94 N

(b) Acceleration of the student = g*sin22.1 - u*g*cos22.1 = 9.81*sin22.1 - 0.123*9.81*cos22.1

= 3.69 - 1.12 = 2.57 m/s^2

(c) Suppose speed at the bottom is v m/s.

so use the expression -

v^2 = u^2 + 2*a*d

put the values -

v^2 = 0 + 2*2.57*19.2

=> v = 9.93 m/s.

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