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insens350 [35]
3 years ago
10

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
Physics
1 answer:
Crank3 years ago
5 0

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

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