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yanalaym [24]
3 years ago
11

A 51.1-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.622 and 0.368,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?
Physics
1 answer:
Charra [1.4K]3 years ago
4 0

Answer:

(a)  The force must be greater than 311.4851 N to just start the crate moving

(b)  The 184.287 N force needed to to slide the crate at constant speed      

Explanation:

We have given that mass of the crate m = 51.1 kg

Coefficient of static friction \mu _s=0.622 and coefficient of kinetic friction \mu _k=0.368

Acceleration due to gravity g=9.8m/sec^2

(A) Static frictional force is given by F_S=\mu _Smg=0.622\times 9.8\times 51.1=311.48516N

So the force must be greater than 311.4851 N to just start the crate moving

(b) For sliding the crate across the dock at a constant speed force must be equal to kinetic friction force

Kinetic friction force is given by F_k=\mu _kmg=0.368\times 9.8\times 51.1=184.287N

So the 184.287 N force needed to to slide the crate at constant speed

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If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
  • force to move the object: F

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
Briefly explain how a resonance tube works​
Lera25 [3.4K]

Answer:

As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency.

4 0
2 years ago
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an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of
Hitman42 [59]

The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v =  Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

Learn more about Concave mirror here:

brainly.com/question/3727024

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3 0
2 years ago
Convert 11 milliseconds into seconds
Alex17521 [72]

Answer:

0.011

Explanation:

6 0
3 years ago
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A 2.0m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the s
Blizzard [7]

Answer:

v=1.92m/s

Explanation:

Given data

Length h=2.0m

Angle α=25°

To find

Speed of bob

Solution

From conservation of energy we know that:

P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\v^{2}=\frac{gh}{0.5}\\ v=\sqrt{\frac{gh}{0.5}}\\ v=\sqrt{\frac{(9.8m/s^{2} )(2.0-2.0Cos(25^{o} ))}{0.5}}\\v=1.92m/s

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3 years ago
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