speed of a block physics question?

Question

A 0.266-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 60 N/m. The block sticks to the spring and the spring compresses 0.14 m before coming to a momentary halt. What is the speed of the block just before it hits the spring?

Answer 1

Using energy
.5*m*v^2+m*g*x=.5*k*x^2
v=sqrt(k*x^2/m-2*g*x)
=sqrt(60*0.14^2/0.266-2*9.81*0.14)
=1.29 m/s

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Answer 2

a. First you will could locate the circumference of the circle the block and string are making. so, 2?R =2(60)? =376.ninety 9 cm If the string is swinging at seventy 5 rounds consistent with minute, it is going seventy 5 x 376.ninety 9 = 28274.33 cm/min in case you elect this in meters/2d (as you will for area b), its 4.712 m/s b. because of the fact the string in swinging in a circle, the acceleration is pointing in the direction of the middle/beginning place of the circle. ?F=ma, and the only rigidity appearing on the stone in this concern is the rigidity T. so, T=ma and the acceleration is centripetal, so the equation reads T=mv²/R you will want the mass in kg, so it somewhat is going to likely be .2 kg the size could be in meters, so it somewhat is going to likely be .6 m in case you replace the variables, you're able to have T =(.2)(4.seventy one²)/.6 this comes out to a grand entire of T = seventy 3.ninety 4 Newtons

Answer 3

♠ coming to a momentary halt the spring has pot energy
E=0.5k*x^2, where x=0.14m, k=60N/m;
♣ meanwhile initial pot energy of the block was E=mg*(h+x), where h is the height above relaxed spring;
energy conserving we get E=E, that is
♦ 0.5k*x^2 = mg*(h+x); hence mgh = 0.5k*x^2 –mgx;
and when the block comes to contact mgh=0.5m*v^2;
0.5k*x^2 –mgx = 0.5m*v^2, hence
v=√(k*x^2 /m –2gx) =
= √(60*0.14^2 /0.266 –2*9.8*0.14) =1.295m/s;

Answer 4

(1/2) m V^2 + m g x = (1/2) k x^2
V^2 = (k/m) x^2 - 2 g x = 1.677 (m/s)^2
V = 1.29 m/s