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luda_lava [24]
3 years ago
10

A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the

rider and his bike is 88.0 kg. Sea level is the zero level for measuring gravitational potential energy. Ignoring friction and air resistance, what is the rider’s total mechanical energy when he coasts to a height of 25.0 m above sea level
Physics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

The mechanical energy of the rider at any height will be 6.34 × 10⁴ J.

Explanation:

Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

Then, calculating the potential and kinetic energy at 55.0 m and 19 m/s, we can obtain the mechanical energy that will be constant:

Mechanical energy = PE + KE

Where:

PE = potential energy.

KE = kinetic energy.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity.

h = height.

Then, the potential energy of the rider will be:

PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" its velocity. Then:

KE = 1/2 · 88.0 kg · (19.0 m/s)²

KE = 1.59 × 10⁴ J

The mechanical energy of the rider will be:

Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J

This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

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Answer:

164.87 J

Explanation:

From the question,

Work done (W) = mghcosθ........................ Equation 1

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Given: m = 25 kg, h = 2.6 meters, θ = 75°.

Constant: g = 9.8 m/s²

Substitute these value into equation 1

W = 25×9.8×2.6×cos75°

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Answer:

Explanation:

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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
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Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

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which yields:

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b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

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We know that V_{0} is zero, so we can simplify the expression:

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V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

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which yields:

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We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

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