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
Serga [27]
3 years ago
7

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

(a) What is the magnitude of the least acceleration the monk"y must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) magnitude and (c) direction of the monkey's acceleration and (d) the tension in the rope?

Physics
1 answer:
pshichka [43]3 years ago
5 0

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

You might be interested in
About how long have wastewater treatment plants been in existence? about how long have wastewater treatment plants been in exist
ella [17]
Water treatment plants have been in existence for almost 220 years
8 0
3 years ago
The slope of a velocity versus time graph gives
Marina86 [1]

Explanation:

Average of acceleration

4 0
3 years ago
I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving
snow_lady [41]

Answer:

9.7\times 10^{-4}\ rad/s

Explanation:

Given:

m=500 kg\\F=3000 N

Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\  know\  that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}

=\frac{6}{6371 \times 10^3\ m}

=9.7\times 10^{-4}\ rad/s

7 0
3 years ago
Use the equation PiVi PfVf. Assume that Pi 101 kPa and Vi 10.0 L. If Pf 43.0 kPa, what is Vf
iren2701 [21]
Solving for vf gives you PiVi/Pf. Now plug in 101kPa*10L/43kPa = 23.48L. Using significant figures i would round to 23.5L
8 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
Other questions:
  • A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introd
    8·1 answer
  • Select the true statement regarding first-order neurons.
    6·1 answer
  • Which best explains why a wood- burning fireplace represents an open system?
    14·2 answers
  • What is the weight on earth of a girl with a mass of 39 kg? N
    10·1 answer
  • Jill tells Bobby that the lightest element in the periodic table is Chlorine because it has an atomic number of 17. Bobby states
    6·1 answer
  • How much work is done if a force of 20 N moves an object a distance of 6 m?​
    11·1 answer
  • Use the excerpts and your knowledge of social studies to answer the question.
    11·2 answers
  • If you could travel 900 meters in 55 seconds, what is your speed?<br> (Answer in details=brainliest)
    6·1 answer
  • If the same total travel at the speed of 0.05 m/s for 40 seconds, how far would it have gone?
    10·1 answer
  • Which of the following is the least important factor of a personal fitness program? A. the individual's personal conditions B. t
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!