A 140 g block is dropped onto a relaxed vertical spring that has a spring consta

Q: A 140 g block is dropped onto a relaxed vertical spring that has a spring consta

a) W = m*g*x = 0.140*9.8*0.18= 0.24696J b) W =-0.5*kx^2 = -05*110*0.18 =-9.9J c) from the work energy thoerem 0.5 m v^2 = ( W1+ W2) - 0.5 mv^2 =( 0.24696-9.9) v = root 2 ( -9.65304)/0.140 =11.74 m/s d) if speed is doubled v' = 2v = 2(11.74) =23.48 m/s from the work energy theorem mgd' - 1/2k d'^2= 1/2 mv'^2 0.140*9.8*d'-0.5*110*d'^2 =-0.5*0.140*23.48^2 55 d'^2-1.372d'-38.59 = 0 solving quadratic equaiton d' = 0.85 m

ID: 3281225 • Letter: A

Question

A 140 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 1.1 N/cm (see the figure). The block becomes attached to the spring and compresses the spring 18 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

a) W = m*g*x = 0.140*9.8*0.18= 0.24696J

b) W =-0.5*kx^2 = -05*110*0.18 =-9.9J

c) from the work energy thoerem

0.5 m v^2 = ( W1+ W2)

- 0.5 mv^2 =( 0.24696-9.9)

v = root 2 ( -9.65304)/0.140

=11.74 m/s

d) if speed is doubled

v' = 2v = 2(11.74) =23.48 m/s

from the work energy theorem

mgd' - 1/2k d'^2= 1/2 mv'^2

0.140*9.8*d'-0.5*110*d'^2 =-0.5*0.140*23.48^2

55 d'^2-1.372d'-38.59 = 0

solving quadratic equaiton d' = 0.85 m

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A 140 g block is dropped onto a relaxed vertical spring that has a spring consta

A 140 g block on a frictionless table is firmly attached to one end of a spring

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A 140 g block is dropped onto a relaxed vertical spring that has a spring consta

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