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A 1.4 kg wood block is launched up a wooden ramp that is inclined at a 22 deg angle. The block's initial speed is 14 m/s. Use Uk = 0.20 for the coefficient of kinetic friction for wood on wood.

U = mu...

(a) What vertical height does the block reach above its starting point?

(b) What speed does it have when it slides back down to its starting point?

Okay so this is how I started it...

I drew a force diagram of the object on an inclined ramp at 22 deg. Set the x axis parallel with the object.. so basically at the same angle. 3 forces.. normal, weight, and kinetic friction.

E = summation

M = mass

A = accel

N = normal

Mg = mass * gravity

fk = kinetic friction

E(Fy) = MAy = 0

N - Mg*cos(22) = 0

N = 12.73

E(Fx) = MAx

-fk - Mg*sin(22) = MAx

-Uk*N - Mg*sin(22) = MAx

-0.20*12.73 - 5.14 = MAx

Ax = -7.686 m/s^2

Vf = Vi + a*t

t = 1.95s

Plug that into the x kinematic equation...

xf = xi + vi*t + (1/2)*a*t^2

= 0 + 15*sin(22)*1.95 + (1/2)*(-7.68)(1.95^2)

= 12.507 m

The answer is wrong.. if someone could disect my work and figure out what I'm doing wrong that'd be awesome, thanks!