# 54. A 190 g block is dropped onto a relaxed vertical spring that has a spring
ID: 1688467 • Letter: #
Question
# 54. A 190 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 1.5 N/cm (Fig. 7-43). The block becomes attached to the spring and compresses the spring 11 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?a is .20482
b is .9075
c is 2.72
and d is!?!?!?!
Explanation / Answer
mass of the block m =190 g=190*10^-3 kgspring constant of k = 1.5 N/cm =1.5*10^2 N/m
compreessing length ?x =11 cm=11*10^-2 m a) work done by gravity W_1 =mg?x =(190*10^-3 kg)(9.8 m/s^2)(11*10^-2 m) W_1 =0.20482 J b) work done by spring force W_2 =-1/2(K?x^2)=-1/2[(1.5*10^2 N/m)(11*10^-2 m)^2] =-0.9075 J c)speed of block(v) : due to conservation of energy ,K.E = W_1+W_2 -1/2(mv^2) =W_1+W_2 -v^2 =2/m(W_1+W_2) -v^2=(2/190*10^-3 kg)(0.20482 J-0.9075 J) v =2.72 m/s d) let speed be doubled i.e., v' =2v due to conservation of energy ,K.E = W'_1+W'_2 -1/2(m(v')^2) =mg?x'-1/2(k(?x')^2) -1/2(m(4v^2)) =mg?x'-1/2(k(?x')^2) -2mv^2 =mg?x'-1/2(k(?x')^2) -2(190*10^-3 kg)(2.72 m/s)^2 =(190*10^-3 kg)(9.8 m/s^2)(?x')-1/2(1.5*10^2 N/m)(?x')^2 75(?x')^2-1.862(?x')-2.811=0 ?x' =1.862
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