1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ZanzabumX [31]
3 years ago
8

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 52.1 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block
Physics
1 answer:
Sophie [7]3 years ago
7 0

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

         F - fr = 0

         F = fr

         fr = 52.1 N

Now we can work  with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

        N - W₁ -W₂ = 0

        N = W₁ + W₂

as the two blocks are identical

        N = 2W

X axis

        F - fr₁ - fr₂ = 0

        F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

          fr₁ = 52.1 N

          fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

          fr₂ = μ 2N

          fr₂ = 2 μ N

          fr₂ = 2 fr₁

we substitute

          F = fr₁ + 2 fr₁

          F = 3 fr₁

          F = 3  52.1

          F = 156.3 N

You might be interested in
For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex
rjkz [21]
Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
 
1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches 

4 0
3 years ago
The formula for speed is v = s/t. <br> a. True<br> b. False
crimeas [40]
If you mean S is the distance then it is true 
Velocity = Distance / time 
3 0
3 years ago
3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
andreev551 [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0
3 years ago
If a 20 N object has been lifted 5 meters above the ground, how much gravitational potential energy does it have?
Andreyy89

The gravitational potential energy of the object is 100 J.

Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.

The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.

Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.

GPE = mgh

The weight of the object is given as 20 J and it is raised to a height of 5 m.

GPE =(mg)*h = (20 N)*(5m)=100 J

The gravitational potential energy of the object is 100 J.

5 0
3 years ago
Read 2 more answers
Any transparent material that allows light of only certain color to pass through is called a filter . true or false
Dahasolnce [82]

it is TRUE hope this helps

7 0
3 years ago
Other questions:
  • A marble rolls off a table with a speed of 2.30 m/s. If the table is 1.12 m high, how far from the
    10·1 answer
  • Is the average speed related to the maximum or minimum speed of the person
    10·2 answers
  • A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (
    8·1 answer
  • Alternating current is normally produced by a
    8·1 answer
  • • Two other reasons for the season are
    9·1 answer
  • Scientific notation for 7890
    14·1 answer
  • Which of the following does not accurately depict a scientific theory?
    10·1 answer
  • Sa fait combien 2×5÷6​
    7·2 answers
  • Pls reply... plss!! help me​
    8·1 answer
  • A machine produces 4,000 J of work in 5 seconds. how much power does the machine produce.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!