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RoseWind [281]
3 years ago
13

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of 180 N. The frictional for

ces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of 20.0 cm and rotates at 2.50 rev/s. The coefficient of kinetic friction between the wheel and the tool is 0.320. At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?
Physics
1 answer:
notka56 [123]3 years ago
8 0

Answer:

P = 57.6 N * 3.14 m/s = 180.956 W

Explanation:

Data given

F = 180 N represent the applied force by the wheel

r = 0.2m represent the radius for the wheel

w= 2.5 rev/s represent the rotational speed given

\mu_k = 0.32 represent the kinetic friction coeffcient given.

Solution to the problem

For this case we need to find first the spped with the usual units in m/s like this:

v = \frac{2\pi R}{T} = 2\pi R w

And we can replace the values given and we got:

v = 2\pi (0.2 m) (2.5 rev/s) = 3.142 m/s

Now we can calculate the friction force, for this case we assume that the normal force is the force exerted by the wheel and from the definition:

F_k = \mu_k F = 0.32*180 N= 57.6 N

Now we can calculate the power assuming that all the energy is converted to kinetic energy from the motor with the following formula:

P = F v

And we can replace with the values that we found:

P = 57.6 N * 3.14 m/s = 180.956 W

And that would be the energytransferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool.

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Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional ela
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Answer:

a) The block 1 slides 0.24 m into the rough region.

b) The block 2 slides 2.7 m

Explanation:

Hi there!

First, let´s find the final velocity of each block. With that velocities, we can calculate the kinetic energy of each block. The kinetic energy of the blocks will be equal to the work done by friction to stop them. From the equation of work, we can calculate the distance traveled by the blocks.

Since the collision is elastic, the momentum and kinetic energy of the system composed of the two blocks is constant.

The momentum of the system is calculated as the sum of the momenta of each block:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

m1 and m2 = mass of blocks 1 and 2 respectively.

v1 and v2 = velocity of blocks 1 and 2 respectively.

v1´ and v2´ = final velocity of blocks 1 and 2 respectively.

Using the data we have, we can solve the eqaution for v1´:

m1 · 3.6 m/s + 0.40 m1 · 0 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s · m1 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s = v1´ + 0.40 v2´

v1´ = 3.6 m/s - 0.40 v2´

The kinetic energy of the system also remains constant:

1/2 m1 · (v1)² + 1/2 m2 · (v2)² = 1/2 m1 · (v1´)² + 1/2 m2 · (v2´)²

Multiply by 2 both sides of the equation:

m1 · (v1)² + m2 · (v2)² = m1 · (v1´)² + m2 · (v2´)²

Let´s replace with the data:

m1 · (3.6 m/s)² + 0.40 m1 · 0 = m1 · (v1´)² + 0.40 m1 (v2´)²

divide by m1:

(3.6 m/s)² = (v1´)² + 0.40 (v2´)²

Replace v1´ = 3.6 m/s - 0.40 v2´

(3.6 m/s)² = (3.6 m/s - 0.40 v2´)² + 0.40 (v2´)²

Let´s solve for v2´:

(3.6 m/s)² = (3.6 m/s)² - 2.88 v2´ + 0.16 (v2´)² + 0.40 (v2´)²

0 = 0.56 (v2´)² - 2.88 v2´

0 = v2´(0.56 v2´ - 2.88)   v2´ = 0 (the initial velocity)

0 = 0.56 v2´ - 2.88

2.88/0.56 = v2´

v2´ = 5.1 m/s

Now let´s calculate v1´:

v1´ = 3.6 m/s - 0.40 v2´

v1´ = 3.6 m/s - 0.40 (5.1 m/s)

v1´ = 1.56 m/s

Now, let´s calculate the final kinetic energy (KE) of each block:

a) Block 1:

KE = 1/2 · m1 · (1.56 m/s)² = m1 · 1.2 m²/s²

The work done by friction is calculated as follows:

W = Fr · s

Where:

Fr = friction force.

s = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction.

And the normal force is calculated in this case as:

N = m1 · g

Where g is the acceleration due to gravity.

Then, the work done by friction will be:

W = m1 · g · μ · s

The kinetic energy of an object is the negative work that must be done on that object to bring it to stop. Then:

m1 · 1.2 m²/s² = m1 · g · μ · s

Solving for s:

s = m1 · 1.2 m²/s²  / m1 · g · μ

s = 1.2 m²/s²/ 9.8 m/s² · 0.50

s = 0.24 m

The block 1 slides 0.24 m into the rough region.

b) For block 2 the kinetic energy will be the following:

KE = 1/2 · 0.4 · m1 · (5.1 m/s)² = m1 · 5.2 m²/s²

The friction force will be:

Fr = 0.4 m1 · g · μ

And the work done will be:

W = 0.4 m1 · g · μ · s

Since W = ΔKE,

Then:

m1 · 5.2 m²/s² = 0.4 m1 · g · μ · s

Solving for s:

5.2 m²/s²/(0.4 · g · μ) = s

s =  5.2 m²/s²/(0.4 · 9.8 m/s² · 0.50)

s = 2.7 m

The block 2 slides 2.7 m

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