Physics Help. Answer please?

[匿名]

2015-12-10 4:54 pm

A 1.6-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. What is the velocity of the block at time t = 0.40 s?
A 1.6-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m. The block is pulled from its equilibrium position at = 0.00 m to a displacement = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal -axis. What is the velocity of the block at time = 0.40 s?
-0.82 m/s
-0.30 m/s
0.00 m/s
0.82 m/s
0.30 m/s

回答 (1)

NCS

2015-12-10 10:56 pm

✔ 最佳答案

Let's model with
x(t) = Acos(ωt)
where A = 0.080 m
and ω = √(k/m) = √(190N/m / 1.6kg) = 10.9 rad/s.
So
x(t) = 0.080m * cos(10.9t) → for t in seconds.
Then
v(t) = x'(t) = 0.080m * 10.9rad/s * -sin(10.9t)
which when t = 0.40 s is
v(0.40) = 0.82 m/s ◄

Hope this helps!

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