# A block sliding on ground where μk = .193 experiences a 14.7 N friction force. What is the mass of the block

Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77

### What is Friction?

According to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab), it is not treated as a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Scientists began piecing together the laws governing friction in the 1400s, according to the book Soil Mechanics(opens in new tab), but because the interactions are so complex.

F=μ*m, n=w which also means n=mg, 14.7=0.193*n, n=76.2, 76.2=m*9.8, m=7.77.

Therefore, Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77.

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7.77

Explanation:

F=μ*m

n=w which also means n=mg

14.7=0.193*n

n=76.2

76.2=m*9.8

m=7.77

## Related Questions

When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

• mass of the box, m = 5 kg
• extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

T = m(g - a) = 0

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

we know that;

mg = kX

5 × 9.81 = k(0.05)

k = 981 N/m

a)

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring  will stretch 60.19 mm with the same box attached as it accelerates upwards

B)

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Most elements in nature exist as mixture of two or more isotopes
t or f

Atomic mass is a value that depends on the distribution of an element's isotopes in nature and the masses of those isotopes. Circle the letter of each sentence that is true about a carbon-12 atom. ... Most elements exist as a mixture of two or more isotopes.

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What is the average kinetic energy of hydrogen atoms on the 5500°C surface of the sun?

Answer: The average kinetic energy of hydrogen atoms is

Explanation:

To calculate the average kinetic energy of the atom, we use the equation:

where,

K = average kinetic energy

k = Boltzmann constant =

T = temperature =

Putting values in above equation, we get:

Hence, the average kinetic energy of hydrogen atoms is

Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

The linear velocity is represented by the following expression:

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

(Eq. 1)

Where:

- Radius of rotation of the particle, measured in meters.

- Angular velocity, measured in radians per second.

- Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

(Eq. 2)

Where:

- Angular displacement, measured in radians.

- Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

(Eq. 3)

From Geometry we must remember that circular arc (), measured in meters, is represented by:

The linear velocity is represented by the following expression:

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

Where,

= Mass of each object

= Initial Velocity of each object

= Final velocity

Replacing we have that,

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

Since there is no initial speed, then

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

Therefore the Kinetic friction coefficient is 0.7105

"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 781.25 N when t = 1.27 s .What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?