Question

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40 m/s at the bottom, how many times higher must the hill be

Answer 1

Answer:

To achieve the velocity of 40 m/sec height will become 4 times  

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$

So total energy at the top of the hill $=mgh+0=mgh$

At the bottom of the hill kinetic energy is equal to $\frac{1}{2}mv^2$ and potential energy will be 0

So total energy at the bottom of the hill is equal to $0+\frac{1}{2}mv^2$

Form energy conservation $mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$, for v = 20 m/sec

$20=\sqrt{2\times 9.8\times h}$

Squaring both side

$19.6h=400$

h = 20.408 m

Now if velocity is 0 m/sec

$40=\sqrt{2gh}$

$19.6h=1600$

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times