Energy conservation of a sliding box

[utsharpie]

3.4kg rock starts at 25 cm on a plane with a 33degree angle. upon reaching the bottom the box slides along the horizontal. friction coeff is
.19. how far does the box slide on the horizontal before coming to a rest?

ok i need to use .5MVi2 + MGy1 = .5MVf2 + MGy2 + Ffrd

first i need to find the velocity of the box as it hits the end of the incline. i also need the length of the incline.

i have Y1=.025 meters
Vf=0
Vi=0
M=3.4kg
g=9.8
theta=33
fr=.19

i keep getting 4.3 meters but its wrong

 

[utsharpie]

ok i found the distance from top of incline to bottom of incline to be .459meters using laws of triangles. so if friction is acting the velocity is gsin33-.19gcos33 and that equals 3.7758

v2=2ad
v2=3.466
v=1.8618

ok so we got Vi at the bottom to be 1.861 and the y1 and y2 at the bottom are both zero and the final V is 0. so is it mv2/2=.19d? and that comes out to 31.01425 meters?no way!

 

[kyle8921]

Based on the given information, we can calculate the velocity of the box as it reaches the bottom of the incline using the equation Vf = √(2gh) where h is the height of the incline. In this case, h = 0.025 meters and g = 9.8 m/s². Therefore, Vf = √(2*9.8*0.025) = 0.7 m/s.

Next, we can use the conservation of energy equation to calculate the distance the box will slide on the horizontal before coming to a rest. The equation is given as 0.5*M*Vf² + M*g*h = μ*M*g*d, where μ is the coefficient of friction and d is the distance the box slides.

Plugging in the values, we get:

0.5*3.4*0.7² + 3.4*9.8*0.025 = 0.19*3.4*9.8*d

Simplifying, we get:

1.176 + 0.833 = 6.626*d

Therefore, d = (1.176+0.833)/6.626 = 0.274 meters.

Therefore, the box will slide 0.274 meters on the horizontal before coming to a rest. This shows that energy conservation is at play in this scenario, as the potential energy of the box at the top of the incline is converted into kinetic energy as it slides down, and then into the work done by friction as it slides on the horizontal.