Physics, 30.11.2019 11:31 twinkle713derp

Adockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-n horizontal force to set it in motion. however, after the crate is in motion, a horizontal force of 55 n is required to keep it moving with a constant speed. find the coefficients of static and kinetic friction between crate and floor. static friction kinetic friction

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Answer from: BlackMilkTea

The coefficients of static and kinetic friction between the crate and floor are;

μ_s = 0.3547

μ_k = 0.2672

1) To find the coefficient of static friction, is given by the formula;

F_s = μ_s*mg

We are given;

Mass; m = 21 kg

Horizontal force to set it to motion; F_h = 73 N

Now, in this static friction equation, we are using this force of 73 N because static friction is friction at rest and 73 N is the force to pull the object from rest. Thus;

73 = μ_s(21 × 9.8)

μ_s = 73/(21 × 9.8)

μ_s = 0.3547

2) To find the coefficient of kinetic friction, is given by the formula;

F_k = μ_k*mg

We will make use of the force of 55N because we are dealing with friction associated with motion and 55N is the force that keeps the object in motion. Thus;

μ_k = F_k/(mg)

μ_k = 55/(21 × 9.8)

μ_k = 0.2672

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Answer from: chmereaustin1

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

$F_s = \mu_s mg$

where

$\mu_s$ is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

F_s=73 N

Using this information into the previous equation, we can find the coefficient of static friction:

$\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355$

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

$F_k = \mu_k mg$ (1)

where

$\mu_k$ is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

F-F_k = ma = 0

And by substituting (1), we can find the value of the coefficient of kinetic friction:

$F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267$

Answer from: Quest

the answer is b, have negligible mass

Answer from: Quest

The objects do not reach terminal velocity there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.

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Adockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal sur...