1/ A 2.00-kg block is pushed against a spring with negligible mass and force con
ID: 2123751 • Letter: 1
Question
1/ A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m. compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0O .style="font-family: 'times new roman'; text-indent: 40px; letter-spacing: 0px; " data-mce-style="font-family: 'times new roman'; text-indent: 40px; letter-spacing: 0px;">style="font-family: 'times new roman'; text-indent: 40px; letter-spacing: 0px; " data-mce-style="font-family: 'times new roman'; text-indent: 40px; letter-spacing: 0px;">style="font-family: 'times new roman'; text-indent: 40px; letter-spacing: 0px; " data-mce-style="font-family: 'times new roman'; text-indent: 40px; letter-spacing: 0px;">style="font-size: medium; ">
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(a) What is the speed of the block as it slides along the horizontal surface after having left the spring? style="letter-spacing: 0px; font-size: medium; ">
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(b) How far does the block travel op the incline before starting to slide back down?style="letter-spacing: 0px; font-size: medium; ">
Explanation / Answer
potential energy of spring = kinetic energy
=> 0.5kx^2 = 0,5mv^2
=> v= sqrt(kx^2/m) = sqrt(400*0.22*0.22/2) = 3.11127 m/sec ans(1)
angle = 37 degrees
thus height = distance*sin(37) = D*0.60182
thus :
mgh = 0.5mv^2
=> 2*9.8*D*0.60182 = 0.5*2*3.11127*3.11127
=> D = 0.82052 answer(2)
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