Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
Kinetic energy: the energy of motion
Work: the change in kinetic energy
Power: the rate of work done
Explanation:
The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where
m is the mass of the object
v is its speed
The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:
![W=K_f - K_i](https://tex.z-dn.net/?f=W%3DK_f%20-%20K_i)
where
is the final kinetic energy
is the initial kinetic energy
Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as
![P=\frac{W}{t}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BW%7D%7Bt%7D)
where
W is the work done
t is the time elapsed
Learn more about kinetic energy, work and power:
brainly.com/question/6536722
brainly.com/question/6763771
brainly.com/question/6443626
brainly.com/question/7956557
#LearnwithBrainly
<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem.
The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So:
0.30 * 2N = 0.6N
And if you look at your options, you'll see that option "B" matches exactly.</span>
Answer:
Hello your question is incomplete attached below is the complete question
Answer : x ( acceleration of mass 4m ) = ![\frac{g}{7}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B7%7D)
The top pulley rotates because it has to keep the center of mass of the system at equilibrium
Explanation:
Given data:
mass suspended = 4 meters
mass suspended at other end = 3 meters
first we have to express the kinetic and potential energy equations
The general kinetic energy of the system can be written as
T = ![\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2](https://tex.z-dn.net/?f=%5Cfrac%7B4m%7D%7B2%7D%20x%5E2%20%20%2B%20%5Cfrac%7B3m%7D%7B2%7D%20%28-x%2By%29%5E2%20%2B%20%5Cfrac%7Bm%7D%7B2%7D%20%28-x-y%29%5E2)
T =
also the general potential energy can be expressed as
U = ![-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant](https://tex.z-dn.net/?f=-4mgx-3mg%28-x%2By%29-mg%28-x-y%29%2Bconstant%3D-2mgy%20%2Bconstant)
The Lagrangian of the problem can now be setup as
![L =4mx^2 +2my^2 -2mxy +2mgy + constant](https://tex.z-dn.net/?f=L%20%3D4mx%5E2%20%2B2my%5E2%20-2mxy%20%2B2mgy%20%2B%20constant)
next we will take the Euler-Lagrange equation for the generalized equations :
Euler-Lagrange equation = ![4x-y =0\\-2y+x +g = 0](https://tex.z-dn.net/?f=4x-y%20%3D0%5C%5C-2y%2Bx%20%2Bg%20%3D%200)
solving the equations simultaneously
x ( acceleration of mass 4m ) = ![\frac{g}{7}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B7%7D)
The top pulley rotates because it has to keep the center of mass of the system at equilibrium
given that
mass of ball = 0.095 kg
initial velocity of ball towards the wall = 40 m/s
final velocity of the ball after it rebound = 30 m/s
now change in momentum is given as
![\Delta P = m(v_f - v_i)](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20m%28v_f%20-%20v_i%29)
![\Delta P = 0.095(30 - (-40))](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%200.095%2830%20-%20%28-40%29%29)
![\Delta P = 6.65 kg m/s](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%206.65%20kg%20m%2Fs)
So change in momentum will be 6.65 kg m/s