a 545 g block is pushed into a spring (k=485N/m) a distance of18.0cm. When the b

Question

a 545 g block is pushed into a spring (k=485N/m) a distance of18.0cm.   When the ball is released what is itsvelocity? The block slides across a smooth surface once it leaves thespring and then up a ramp.   It travels up the ramp adistance of 275 cm What is the elevation angle of theramp? a 545 g block is pushed into a spring (k=485N/m) a distance of18.0cm.   When the ball is released what is itsvelocity? The block slides across a smooth surface once it leaves thespring and then up a ramp.   It travels up the ramp adistance of 275 cm What is the elevation angle of theramp?

Explanation / Answer

mass m = 545 g = 0.545 kg spring constant k = 485 N / m compression x = 18 cm = 0.18 m At release point , kinetic energy = maximum elasticpotential energy      ( 1/ 2) m v^ 2 = ( 1/ 2) k x ^2 from this v = x [ k / m ]                  = 5.369 m / s (b).Initial velocity v = 5.369 m / s final velocity V = 0 distance S = 275 cm = 2.75 m Accleration a = - g sin from the relation V ^ 2 - v ^ 2 = 2aS                                   - v ^ 2 = -2gS sin from this sin = v ^ 2 / 2gS                       = 0.534                     = 32.33 degree

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