3 years ago

I didnt want my question public i made a mistake i want it taken down

Physics

1 answer:

mrs_skeptik [129]3 years ago

5 0

Then report it and it might be taken down

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3 years ago

Read 2 more answers

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 c

Bas_tet [7]

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

$x = \sqrt{\frac{mv_0^2}{k} } \\\\x = \sqrt{\frac{1.6(0.95)^2}{902} }\\\\x = 0.040 \ m$

x = 4 cm

(b) speed of the block

Apply the law of conservation of energy;

¹/₂mv² = ¹/₂kx²

mv² = kx²

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

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3 years ago