A 5.00-kg block slides on an inclined plane. The incline is 40.0 degree above th

Question

A 5.00-kg block slides on an inclined plane. The incline is 40.0 degree above the horizontal. The block is released from rest and reaches a speed of 2.50 m/s at the bottom. What is the distance up the incline that the block starts from? Use conservation of energy principle to solve the problem x = vt; x = V_o t + 1/2 at^2 V_f = V_o + at y = y_o + V_o t + 1/2 at^2 V = V_f + V_o/2 V^2_f = V^2_o + 2ax g = 9.8 m/s^2 F_net = ma f = mu N W = mW = FS or W = FS Cos theta; P = Work/t or Energy/time; P = FV G.P.E = mgh; K.E = mV^2/2 W_net = Delta K.E; W_ne = Delta G.P.E + Delta K.E + In the presence of non conservative forces Delta G.P.E + Delta K.E = 0 (In the absence of non-conservative forces)

Explanation / Answer

using energy conservation:

KEi + PEi = KEf + PEf

KEi = 0 since block is released with zero velocity

PEf = 0 w.r.t ground

m*g*h = 0.5*mv^2

h = v^2/(2*g)

h = 2.5^2/(2*9.81)

h = 0.318 m

distance up the incline will be

sin A = h/D

D = h/sin 40 deg

D = 0.318/sin 40 deg = 0.494 m

0.494 m up the incline

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