Otto the Otter (mass = 7.5 kg) takes a trip down his water slide, starting from

Question

Otto the Otter (mass = 7.5 kg) takes a trip down his water slide, starting from a stationary position. The slide has a length of 4 cm and is angled at 40 degrees with respect to horizontal. The end of the slide is parallel to the surface and 2.5 m above a pond. The otter-water coefficient of kinetic friction is mu _k = 0.10. Otto's aerodynamic body minimizes drag force as he files through the air. Using a free body diagram and your knowledge of newton's laws and common forces, show that the normal force on Otto is N = 56.3 N and that the component of Otto's weight directed down the ramp is W_D = 47.2 N. Determine the frictional force on Otto as he sides down the ramp. Show that Otto's speed as he leaves the ramp is v_0 = 6.7 m/s.

Explanation / Answer

part a:

normal force=mass*g*cos(theta)

=7.5*9.8*cos(40)=56.3043 N=56.3 N (approx)

component of weight down the ramp=weight*sin(40)

=7.5*9.8*sin(40)=47.245 N=47.2 N(approx)

part b:
friction force=friction coefficient*normal force

=0.1*56.3043=5.63043 N

part c:

net force=force along the inclince-friction force

=47.245-5.63043=41.6146 N

net acceleration=net force/mass=41.6146/7.5=5.5486 m/s^2

distance covered=4 m

initial velocity=0 m/s

then final velocity=sqrt(initial velocity^2+2*acceleration*distance)

=sqrt(0+2*5.5486*4)=6.6625 m/s

so speed at which the otter flies off the slide is 6.6625 m/s which is approximately 6.7 m/s

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