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

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

Step-by-step 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
`(1)/(2)* 70* 12^2+70* 9.8* 30=12* 450+0+(1)/(2)* 70* v^2`

`(1)/(2)* 70* v^2=20220`

`v=24.03m/sec`

Answer 2

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

Step-by-step 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

`(1)/(2)mv^2+mgh=Fx+(1)/(2)mv^(2)`

`(1)/(2)*70*12^2+70*9.8*30=12*450+(1)/(2)*70*v^2`

`25620=5400+35v^2`

`35v^2=25620-5400`

`35v^2=20220`

`v^2=(20220)/(35)`

`v=\sqrt{(20220)/(35)}`

`v=24.035\ m/s`

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