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Ilia_Sergeevich [38]
3 years ago
15

A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

from rest and is on a level road, what speed ???? will he be going after 35.0 m? The mass of the bicyclist and his bicycle is 90.0 kg.

Physics
1 answer:
nalin [4]3 years ago
8 0

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

v_{f}^{2} = v_{i}^{2} + 2ad        

v_{f} = \sqrt{v_{i}^{2} + 2ad}  (1)

The acceleration can be found by means of Newton's second law:

\sum F_{net} = ma

Where \sum F_{net} is the net force, m is the mass and a is the acceleration.

Fx + Fy = ma  (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

F_{x} = F - F_{air}  (3)

Forces in the y axis:

F_{y} = 0 (4)

Solving for the forces in the x axis:

F_{x} = F - F_{air}

Where F = 1.150x10^{3} N and F_{air} = 795 N:

F_{x} = 1.150x10^{3} N - 795 N

F_{x} = 355 N

Replacing in equation (2) it is gotten:

Fx + Fy = ma

355 N + 0 N = (90.0 Kg)a

355 N = (90.0 Kg)a

a = \frac{355 N}{90.0Kg}

a = \frac{355 Kg.m/s^{2}}{90.0Kg}

a = 3.94 m/s^{2}

So the acceleration for the cyclist is 3.94 m/s^{2}, now that the acceleration is known, equation (1) can be used:

v_{f} = \sqrt{v_{i}^{2} + 2ad}

However, since he was originally at rest its initial velocity will be zero (v_{i} = 0).

v_{f} = \sqrt{2ad}

v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}

v_{f} = 16.60m/s

He has a speed of 16.60m/s after 35.0 meters

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