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stepladder [879]
3 years ago
7

A rollercoaster car has 2500 J of potential energy and 160 J of

Physics
1 answer:
lara31 [8.8K]3 years ago
8 0

Answer:

E_{k2}=2660 [J] kinetic energy.

Explanation:

The energy in the initial state i.e. when the rollercoaster is at the top is equal to the energy in the final state i.e. when it is at the bottom of the hill.

These states can be represented by means of the second equation.

E_{k1}+E_{p1}=E_{k2}\\160 + 2500 = E_{k2}\\E_{k2}=2660 [J]

Since the rollercoaster is located in the bottom of the hill where the potential energy level is zero, therefore there is only kinetic energy in the second state.

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Xavier is roller skating at 14 km/h and tosses a set of keys forward on the ground at 8 km/h. The speed of the keys relative to
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22 km/h

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

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(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

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W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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