Question

At sea world, a 900 kg polar bear slides down a wet slide inclined at an angle of 25.0 to the horizontal. The coefficient of friction between the bear and the slide is 0.0500. What frictional force impedes the bears motion down the slide?

Answer 1

The force of friction is 399.7 N

Explanation:

The force of friction acting on the bear is given by

$F_f=\mu N$ (1)

where

$\mu=0.05$ is the coefficient of friction

N is the normal reaction exerted by the plane on the bear

The normal reaction can be  found by analyzing the forces acting in the direction perpendicular to the plane: in fact, the bear is in equilibrium along this direction, so the normal reaction must balance the component of the weight perpendicular to the plane:

$N=mg cos \theta$

where:

m = 900 kg is the mass of the bear

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=25^{\circ}$ is the angle of the incline

Substituting everything into (1), we find the force of friction:

$F=(0.05)(900)(9.8)(cos 25)=399.7 N$

Learn more about inclined plane and frictional force here:

brainly.com/question/5884009

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