Question

A bicyclist starting from rest applies a force of 346N to ride his bicycle across flat ground for a distance of 270m before encountering a hill making an angle of 29 degrees with respect to the horizontal. The bicycle and rider have a mass of 132kg combined. In this problem you can ignore air resistance and other losses did to friction. What is the bicycle’s speed, v in m/s just before the hill?

Answer 1

We are asked to determine the velocity of a bicycle when a force of 346 N is applied to it. To do that we will use a balance of energy.

The work applied to the bike is equivalent to the change in kinetic energy of the bicycle. Therefore, we have:

$W=\frac{1}{2}mv^2_f-\frac{1}{2}mv_0^2$

since the bicycle starts from rest this means that the initial velocity is zero:

$W=\frac{1}{2}mv_f^2$

Since work is the product of force by distance we have:

$Fd=\frac{1}{2}mv^2$

Where "F" is the force and "d" is the distance.

Now, we solve for the velocity. To do that we will multiply both sides by 2:

$2Fd=mv^2$

Now, we divide both sides by the mass:

$\frac{2Fd}{m}=v^2$

Now, we take the square root to both sides:

$\sqrt{\frac{2Fd}{m}}=v$

Now, we substitute the values:

$\sqrt{\frac{2(346N)(270m)}{132kg}}=v$

Solving the operations:

$37.6\frac{m}{s}=v$

Therefore, the velocity is 37.6 meters per second.