A 190 g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 190 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.9 N/cm (see the figure). The block becomes attached to the spring and compresses the spring 16 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

mass of the block m=190g =190*10^-3 kg

spring constant k=2.9 N/cm =290 N/m

compression x=16cm =0.16 m

a)

work done by the gravitational force is,

w1=m*g*x

w1=190*10^-3*9.8*0.16

w1=0.29 J

b)

work done by the spring force is

w2=1/2*k*x^2

w2=1/2*290*0.16^2

w2=3.71 J

c)

by using work-energy theorem,

U1+K1=U2+K2

w1+1/2*m*v1^2=w2

1/2*m*v1^2=(w2-w1)

1/2*190*10^-3*V1^2=(3.71-0.29)

==>

speed of the block just before its hits the spring is,

v1=6 m/sec

d)

let

compression is x2 if speed v1=12 m/sec

then,

w1+1/2*m*v1^2=w2

(m*g*x2)+(1/2*m*12^2)=(1/2*k*x2^2)

(190*10^-3*9.8*x2)+(1/2*190*10^-3*12^2)=(1/2*290*x2^2)

===>

x2=0.314 m

x2=31.4 cm

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