Question

A student of mass 60.0 kg, starting at rest, slides down a slide 20.0 m long, tilted at an angle of 30.0° with respect to the horizontal. If the coefficient of kinetic friction between the student and the slide is 0.120, find (a) the force of kinetic friction, (b) the acceleration, and (c) the speed she is traveling when she reaches the bottom of the slide

Answer 1

Answer:

Explanation:

Reaction force of inclined surface

R = mgcos30

=60 x 9.8 x .866

= 509 N

Force of kinetic friction

= .12 x 509

= 61.08 N

b )

net force downwards = mgsinθ - μmgcosθ

acceleration = g sinθ - μgcosθ

9.8 sin30 - .12 x 9.8 x cos30

= 4.9 - 1.018

a= 3.88 m /s²

c )

v² = 2as

= 2 x  3.88 x 20

v = 12.46 m /s