A package of mass 22 KG slides down a ramp 4.2 m Lon and inclined at an angle if

Question

A package of mass 22 KG slides down a ramp 4.2 m Lon and inclined at an angle if 36 degrees below the horizontal.
A.) what is the mechanical energy at the top if the ramp, If the package has an initial velocity if 4.3 m/s? B. What is the energy lost due to friction as the package slides a distance of 1.2 down the ramp. If Us = .45 and Uk = .32 C. What is the velocity if the package at this position? A package of mass 22 KG slides down a ramp 4.2 m Lon and inclined at an angle if 36 degrees below the horizontal.
A.) what is the mechanical energy at the top if the ramp, If the package has an initial velocity if 4.3 m/s? B. What is the energy lost due to friction as the package slides a distance of 1.2 down the ramp. If Us = .45 and Uk = .32 C. What is the velocity if the package at this position?
A.) what is the mechanical energy at the top if the ramp, If the package has an initial velocity if 4.3 m/s? B. What is the energy lost due to friction as the package slides a distance of 1.2 down the ramp. If Us = .45 and Uk = .32 C. What is the velocity if the package at this position?

Explanation / Answer

a) KE = 0.5mv^2

KE = 0.5*22*4.3^2 = 203.39 J

b) Work done = friction*d

W = u*mgcos(theta)*d

W = 0.32*22*9.8*cos36*1.2

W = 66.9788 J

c) KEi = KEf + W

KEf = KEi - W = 203.39 - 66.9788

0.5*mv^2 = 136.411

0.5*22*v^2 = 134.411

v^2 = 12.4

v = 3.5215 m/s

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