Need help with physics problem!! please help please! thanks!

Question

a 50 lb package is projected up a 20 degree incline with an initial velocity of 40 ft/s. knowing that the coefficient of kinetic friction between the package and the incline is .15, determine (a)the maximum distance x that the package will move up the incline, (b)the velocity of the pacage as it returns to its original position, (c) the total amount of energy dissipated due to friction. please help, thank you so much!!

Answer 1

Solve this using conservation of energy:  The initial kinetic energy of the block is 0.5*m*v0²;  when it stops on the incline, that kinetic energy has been converted to gravitational potential energy = m*g*h, where h is the vertical distance achieved, plus heat energy from friction, which is Fr*x, where Fr is the frictional force and x the distance along the ramp.

0.5*m*v0² = m*g*h + Fr*x

h = x*sinθ   Fr =µ*m*g*cosθ

substituting those values get

0.5*m*v0² = m*g*x*sinθ + µ*m*g*x*cosθ

x*g*(sinθ  +  Ã‚µ*cosθ) = 0.5*V0²

x =  0.5*V0² / g*(sinθ + µ*cosθ)

It is easier to get part (c) first, since the energy lost to friction on the way up is Fr*x and on the way down is also Fr*x, so total energy lost to friction is Ef = 2*Fr*x

Ef =  2*µ*m*g*x*cosθ

The velocity on return to the original position (V1) is found from the kinetic energy at return = 0.5*m*V1².  This equals the initial kinetic energy less that lost to friction, so

0.5*m*V1² = 0.5*m*V0² - Ef

0.5*m*V1² = 0.5*m*V0² - 2*µ*m*g*x*cosθ

V1 = √[V0² - 4*µ*g*x*cosθ]

Since x = 0.5*V0² / g*(sinθ + µ*cosθ)

V1 = V0√[1 - 2*µ*cosθ / (sinθ + µ*cosθ)]