Favorites Tools Help -3 points SerCP11 4.P 022 A student of mass 55.4 kg, starti
ID: 1872738 • Letter: F
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Favorites Tools Help -3 points SerCP11 4.P 022 A student of mass 55.4 kg, starting at rest, slides down a slide 20.2 m student and the slide is 0.103, find the force of kinetic friction, the acceleration, and the speed she is traveling when she reaches the bottom of the slide. (Enter the magatudes) ong, tilted at an angle of 24.1 with respect to the horkzontal. If the coeflidient of kinetic friction between the HINT (a) the force of kinetic friction (in N) (b) the acceleration (in m/s') m/s (c) the speed she is traveling (in m/s) m/s Need Help? ResdWach Submit Answer Save Progress Submit Assignment SaExplanation / Answer
a) Normal force acting on the person, N = m*g*cos(theta)
= 55.4*9.8*cos(24.1)
= 495.6 N
kinetic friction, fk = mue_k*N
= 0.103*495.6
= 51.0 N
b) acceleration of the person, a = Fnet/m
= (m*g*sin(theta) - fk)/m
= (55.4*9.8*sin(24.1) - 51.0)/55.4
= 3.08 m/s^2
c) use, vf^2 - vi^2 = 2*a*d
vf^2 - 0 ^2 = 2*3.08*20.3
==> vf = sqrt(2*3.08*20.3)
= 11.2 m/s
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