Okay guys i need some help on this problem: A 775kg meteorite buries itself 5.0

Question

Okay guys i need some help on this problem:

A 775kg meteorite buries itself 5.0 meters into soft mud. The trajectory of the meteorite is vertical, paving its way through the mud until it finally comes to rest 5 meters below the mud -ground level. The magnitude of the force between the meteorite and the mud is given by F(y)= | (640N/m^3)(y^3)| where y is the depth in the mud.

Through the application of the work energy theorem determine the meteriorite's initial speed just before impacting into the mud. Be sure to specify which location you decide to take the gravitational PE level=0 in solving this probelm

Explanation / Answer

dA=F*dy=640*y^3*dy. so A=640*delta(y^4)/4=640*5^4/4=10^5(J). so 10^5=mv^2/2 so v=16.1(m/s)

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---