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
Nadusha1986 [10]
3 years ago
15

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a

nd the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Physics
1 answer:
aleksklad [387]3 years ago
3 0

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

You might be interested in
Question 2 of 20
alukav5142 [94]

Answer: a

Explanation: because i said so

8 0
2 years ago
You leave your home at 1pm. At 3pm, you are 100 km east of your house. What was your average velocity in km/hr
Rasek [7]

Answer:

50km/h

Explanation:

Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)

The change in distance has been given as 100km.

The change in time is 3pm-1pm = 2 hours.

Therefore the average velocity was 100/2 = 50km/h (to the east).

Hope this helped!

4 0
2 years ago
H. Briefly describe the process taking place in the image below and comment on the
katrin [286]

AHEMHFgjrhfrshfghesgrgregr

3 0
3 years ago
Is mass is 88.2 and height of distance is 25m, how do you find speed?
Natasha_Volkova [10]

Answer:

The velocity will be v = 22.1[m/s]

Explanation:

We can solve this problem by using the principle of energy conservation, where potential energy is converted to kinetic energy. For this problem we will take the point with maximum potential energy when the body is 25 [m] high. By the time the height is zero, the potential energy will have been transformed into kinetic energy, and we can find the velocity of the body.

Ep = m*g*h\\where:\\m = mass = 88.2[kg]\\h = elevation = 25[m]\\g = gravity = 9.81 [m/s^2]\\Ep = 88.2*25*9.81 = 21631.05[J]\\

Now we know that the energy will be transformed.

Ek=Ep\\Ek=0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{Ek}{0.5*m} } \\v=\sqrt{\frac{21631.05}{0.5*88.2} } \\v=22.14[m/s]

4 0
2 years ago
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

8 0
3 years ago
Other questions:
  • In each case, you should demonstrate how you worked out your answer, as well as giving the answer.
    15·2 answers
  • What is the main source of energy
    6·2 answers
  • "it was so cold yesterday that the temperature only reached 275." which temperature scale is being used? what would be the corre
    12·2 answers
  • On a dry day, just after washing your hair to remove natural oils and drying it thoroughly, run a plastic comb through it. Small
    15·1 answer
  • Forces affect motions in living and nonliving things. In a human, swallowed food moves down the esophagus into the stomach, even
    10·1 answer
  • A car is stationary at first.it moves off with.acceleration of 2m.s-2.Calculate how far it will move in 10s
    6·1 answer
  • Why are some of the rocks on the moon older than the oldest known rocks on Earth?
    10·2 answers
  • Difference between kinetic energy and potential energy(please i need answer in 2minutes​
    8·1 answer
  • LAST ONE! ASAP PLEASE
    6·2 answers
  • A child jumps on a trampoline, which of the following causes the child to rise in the air?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!