I cant read it, i could most likely help if i could read it.
It's false i hope this helps :)
Answer:
The velocity of other mass is 3.60 m/s.
Explanation:
Given that,
Mass of first block = 8 kg
mass of second block = 4.3 kg
Speed = 6.7 m/s
We need to calculate the speed of first mass
Using conservation of momentum
![(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Bm_%7B2%7D%29u%3Dm_%7B1%7Dv_%7B1%7D%2Bm_%7B2%7Dv_%7B2%7D)
where, m₁ =mass of first block
m₂ =mass of second block
m₁ =mass of first block
v₂ =speed of second block
Put the value into the formula
![8+4.3\times0=8\times v_{1}+4.3\times6.7](https://tex.z-dn.net/?f=8%2B4.3%5Ctimes0%3D8%5Ctimes%20v_%7B1%7D%2B4.3%5Ctimes6.7)
![v_{1}=\dfrac{-4.3\times6.7}{8}](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Cdfrac%7B-4.3%5Ctimes6.7%7D%7B8%7D)
![v_{1}=-3.60\ m/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D-3.60%5C%20m%2Fs)
Negative sign represent the opposite direction of initial value.
Hence, The velocity of other mass is 3.60 m/s.
Given :
Mass of box , m = 250 kg.
Force applied , F = 285 N.
The value of the incline angle is 30°.
the coefficient of dynamic friction is
.
To Find :
The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.
Solution :
Net force applied in box is :
![F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N](https://tex.z-dn.net/?f=F%3D285%20-%20mgsin%5C%20%5Ctheta%20-%20%5Cmu%20mg%20cos%20%5C%20%5Ctheta%5C%5C%20%5C%5CF%3D285-mg%28%20sin%20%5C%20%5Ctheta%20-%20%5Cmu%20cos%5C%20%5Ctheta%29%5C%5C%5C%5CF%3D285%20-%2020%5Ctimes%2010%28%20%5Cdfrac%7B1%7D%7B2%7D%2B0.72%5Ctimes%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%29%5C%5C%5C%5CF%3D60.29%5C%20N)
Acceleration ,
.
By equation of motion :
![v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5C%5Cv%3D0%2B3.01%5Ctimes%204%5C%5C%5C%5Cv%3D12.04%5C%20m%2Fs)
Therefore, the speed of box is 12.04 m/s.
Hence, this is the required solution.
Answer:
Work done in all the three cases will be the same.
Explanation:
1) The free falling body has only one force acting on it, the gravitational force. The work done on the body = mgH (Gravitational potential energy)
2) There are two forces acting on the body going down on a frictionless inclined plane - gravity and the normal force. The gravitational potential energy will be the same. The work done due to the normal force is zero, since the direction of the force is perpendicular to the displacement. Hence, total work done on the body = mgH
3) In the case of the body swinging on the end of a string, the change in gravitational potential enrgy will once again be the same since difference in height is H. The additional force on the body is the tension due to the string. But the work done due to this force is <em>zero, </em>since the displacement of the body is perpendicular to the tension. Therefore, the total work done on the body is once again mgH.