(A) The compression of spring be "4 cm".
(B) Block moving at the speed of "0.427 m/s".
Law of conservation of energy:
According to the question,
Block's mass = 1.6 kg
Spring constant = 902 N/m
Block's initial speed = 0.95 m/s
(A) By applying the law of conservation of energy,
→ [tex]\frac{1}{2}kx^2 = \frac{1}{2}mv_0^2[/tex]
or,
→ [tex]x = \sqrt{\frac{mv_0^2}{k} }[/tex]
By substituting the values,
[tex]= \sqrt{\frac{1.6(0.95)^2}{902} }[/tex]
[tex]= 0.040 \ m[/tex] or, [tex]4 \ cm[/tex]
(B) The speed of block will be:
→ [tex]\frac{1}{2} mv^2 = \frac{1}{2} kx^2[/tex]
or,
→ [tex]v = \sqrt{\frac{kx^2}{m} }[/tex]
[tex]= \sqrt{\frac{(902)(0.018)^2}{1.6} }[/tex]
[tex]= 0.427 \ m/s[/tex]
Thus the above answer is correct.
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