A 1.6 kg block slides with a speed of .95 m/s on a frictionless, horizontal surface until it encounters a spring with a spring constant of 902 N/m.

A) How much does the spring compress when the block comes to rest?

B) When the block has compressed the spring 1.8 cm how fast is the block moving?

A 16 kg block slides with a speed of 95 ms on a frictionless horizontal surface until it encounters a spring with a spring constant of 902 Nm A How much does th class=

Respuesta :

Answer:

(a) the compression of the spring when the block comes to rest is 4cm

(b) speed of the block is 0.427 m/s

Explanation:

Given;

mass of the block, m = 1.6 kg

spring constant, k = 902 N/m

initial speed of the block, v₀ = 0.95 m/s

(a) the compression of the spring when the block comes to rest

when the block comes to rest, the final speed, vf = 0

Apply the law of conservation of energy;

¹/₂kx² = ¹/₂mv₀²

kx² = mv₀²

[tex]x = \sqrt{\frac{mv_0^2}{k} } \\\\x = \sqrt{\frac{1.6(0.95)^2}{902} }\\\\x = 0.040 \ m[/tex]

x = 4 cm

(b)  speed of the block

Apply the law of conservation of energy;

¹/₂mv² = ¹/₂kx²

mv² = kx²

[tex]v = \sqrt{\frac{kx^2}{m}}\\\\v = \sqrt{\frac{(902)(0.018)^2}{1.6}}\\\\v = 0.427 \ m/s[/tex]

(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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