A 300-kg bear grasping a vertical tree slides down at constant velocity. the friction force between the tree and the bear is:___
1 answer:
3000 N of friction exists between the bear and the tree.
The bear goes down at a constant speed, thus there is no acceleration. The forces add up to zero. As a result, upward friction and downward weight result in zero
i.e.,
F-mg=0
F = mg
F = (300 kg)/10 (gravitational acceleration)
3000 N frictional force equals F.
How does friction force work?
- Friction is the force that prevents two solid objects from rolling or sliding over one another. Even though frictional forces, such as the traction required to walk without slipping, may be advantageous, they can also be a significant hindrance to motion.
Between solid surfaces, there are three basic types of friction:
- Rolling
- Sliding.
- Static. They are graded from strongest to weakest. Fluid friction happens when liquids or gasses are mixed together.
To learn more about Friction force, visit:
#SPJ4
You might be interested in
Answer:
terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
Explanation:
Given the data in the question;
we know that, the force on a body due to gravity is;
= mg
where m is mass and g is acceleration due to gravity
Force of drag is;
=
pCAv²
where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.
Terminal velocity is reach when the force of gravity is equal to the force of drag.
mg = pCAv²
we solve for v
v = √( 2mg / pCA )
so we substitute in our values
v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )
v = √( 1685.6 / 0.122015 )
v = √( 13814.6949 )
v = 117.54 m/s
v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr
Therefore terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr