Answer:
The required work done is 
Explanation:
Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '
' be the coefficient of friction, then
where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.
Since the application of force by the movers does not create any acceleration to the block, we can write
So the work done (W) in moving the crate by a distance s = 10.6 m is
Explanation:
Given that,
Mass of a freight car, 
Speed of a freight car, 
Mass of a scrap metal, 
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :
So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
![\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5B%28m_1%2Bm_2%29V%5E2-m_1u_1%5E2%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%2830%2C000%2B110%2C000%20%290.182%5E2-30000%280.85%29%5E2%5D%5C%5C%5C%5C%3D-8518.82%5C%20J)
Lost in kinetic energy is 8518.82. Negative sign shows loss.
Answer: 10.36m/s
How? Just divide 200m by 19.3 and you will get how fast he ran per m/s
The point in which it originates.
Angular momemtum : mass * tangential speed * distance to the center = 50*2.1*3.6=37800 J.s
