Question

A cyclist is coasting at 12 m/s when she starts down a 450-m-long slope that is 30 m high. The cyclist and her bicycle have a combined mass of 70 kg. A steady 12 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 speed of her at the bottom is 24.035 m/s.

Explanation:

Given that,

Speed of cyclist = 12 m/s

Height = 30 m

Distance d = 450 m

Mass of cyclist and bicycle =70 kg

Drag force = 12 N

We need to calculate the speed at the bottom

Using conservation of energy

K.E+P.E=drag force+K.E+P.E

Potential energy is zero at the bottom.

K.E+P.E=drag force+K.E

$\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}$

$\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2$

$25620=5400+35v^2$

$35v^2=25620-5400$

$35v^2=20220$

$v^2=\dfrac{20220}{35}$

$v=\sqrt{\dfrac{20220}{35}}$

$v=24.035\ m/s$

Hence, The speed of her at the bottom is 24.035 m/s.

Answer 2

Answer:

Speed at the bottom of the slope is 24.03 m /sec

Explanation:

We have given speed of the cyclist v = 12 m/sec

She starts down a 450 m long that is 30 m high

Si distance = 450 m and height h = 30 m

Combined mass of cyclist and bicycle m = 70 kg

Drag force = 12 N

We have to find the speed at the bottom

According to conservation theory

KE +PE = work done by drag force + KE + PE

As we know that at the bottom there will be no potential energy

So $\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2$

$\frac{1}{2}\times 70\times v^2=20220$

$v=24.03m/sec$