A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is 96.0 kg, and the hill is
inclined at 17.0° with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?
1 answer:
Answer:
The force applied 275 N in a direction parallel to the hill
Explanation:
Newton's second law is adequate to work this problem, in the annex we can see a free body diagram, where the weight (W) is vertical, the friction force (fr) is parallel to the surface and the normal (N ) is perpendicular to it. In general for these problems a reference system is taken that is parallel to the surface and the Y axis is perpendicular to it.
Let us decompose the weight into its two components, the angle T is taken from the axis and
Wx = W sin θ
Wy = W cos T
We write Newton's second law
∑ F = m a
X axis
The cyclist falls at a constant speed, which implies that the acceleration is zero
fr - W sin θ = 0
fr = mg sin θ
fr = 96 9.8 without 17
fr = 275 N
When the cyclist returns to climb the hill, he must apply the same force he has to overcome the friction force that always opposes the movement . The force applied 275 N in a direction parallel to the hill
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Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
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here
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