A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 26 angle.

Question

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 26 angle. The block's initial speed is 5.0 m/s . The coefficient of kinetic friction of wood on wood is k=0.200. Part A What vertical height does the block reach above its starting point? Express your answer to two significant figures and include the appropriate units. h= Part B What speed does it have when it slides back down to its starting point? Express your answer to two significant figures and include the appropriate units

Explanation / Answer

m = 2kg, u = 5 m/s, theta = 26 degrees , uk = 0.2

Net force along verticla direction

N = mgcos(26)

Fk = uk*N

Net force along horizontla direciton

ma = -mgsin(26) - uk*mg*cos(26)

a = -9.8(sin(26)+0.2*cos(26))

a = -6.06 m/s^2

(a) When the block reaches its maximum height, it s velocity is zero (v = 0), while initial velocity, v equals 10m/s. So find the distance it moves parallel to the incline, and then use trig to find the vertical height it achieves. The parallel distance is:

from kinematic equations

v^2 -u^2 = 2as

0 - 5^2 = -2*6.06*s

s = 2.063 m

Vertical height sin(theta) = h/s

h = 2.063*sin(26)

h= 0.90 m

(b) The acceleration changes for the block as it slides down the incline because now gravity works opposite of friction:

Net force along horizontal direction

ma = mgsin(26)- uk*mgcos(26)

a = 9.8(sin(26) - 0.2*cos(26))

a = 2.53 m/s^2

from kinematic equaitons

v^2 - u^2 = 2as

v^2 -0 = 2*2.53*2.063

v= 3.2 m/s

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