Question

A cyclist is coasting at 13 m/s when she starts down a 460 m long slope that is 30 m high. The cyclist and her bicycle have a combined mass of 70 kg. A steady 15 N drag force due to air resistance acts on her as she coasts all the way to the bottom. What is her speed at the bottom of the slope?

Answer 1

Answer:

The value is  $v = 23.6 \ m/s$

Explanation:

From the question we are told that

   The velocity is $u = 13 \ m/s$

    The  length of the slope is $l = 460 \ m$

    The height of the slope is $h = 30 \ m$

     The mass of the cyclist and her bicycle is $m = 70 \ kg$

   The drag force is  $F = 15 \ N$

Generally from the law of energy conservation we have that

      $PE + KE_t = W * KE_b$

Here $PE$ is the potential energy at the top of the slope which is mathematically represented as

       $PE = m* g * h$

=>     $PE = 70 * 9.8 * 30$  

=>     $PE = 20580 \ J$

and   $KE_t$ is the kinetic energy at the top of the slope which is mathematically represented as

        $KE_t = \frac{1}{2} * m * u^2$

=>     $KE_t = \frac{1}{2} * 70 * 13^2$

=>     $KE_t = 5915 \ J$

And  W is the work done by the bicycle which is mathematically represented as

       $W = F * l$

=>     $W = 15 * 460$

=>     $W = 6900 \ J$

and $KE_b$ is the kinetic energy at the bottom of the slope which is mathematically represented as

       $KE_b = \frac{1}{2} * m * v^2$

=>    $KE_b = \frac{1}{2} * 70 * v^2$

=>    $KE_b = 35 v^2$

So

       $20580 + 5915 = 6900 + 35 v^2$

=>    $v = 23.6 \ m/s$