Question

1. A 3.0-kg block (A) is attached to a 1.0-kg block (B) by a massless spring that is
compressed and locked in place, as shown in figures. The blocks slide without friction along
the x-direction at an initial constant speed of 2.0 m/s. At time t=0, the positions of blocks A
and B are x = 1.0 m and x = 1.2 m, respectively, at which point a mechanism releases the
spring, and the blocks begin to oscillate as they slide. If 2.0 s later block B is located at x =
6.0 m, where will block A be located?
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Answer 1

Answer:

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Answer 2

The answer is x = 5.6 m

Given:

1. A 3.0-kg block

(A) is attached to a 1.0-kg block

(B) by a massless spring that is

the x-direction at an initial constant speed of 2.0 m/s.

To find:

To find the value of X

Explanation:

because block A has a velocity of v = 0.5 m/s

when it reaches x = 6.0 m.

The motion of block A is described by the equation x = a + bt,

where a and b are constants and t is the time.

To find the value of a and b,

we can use the fact that the position of block A is proportional to its velocity and the time,

so we can solve for a and b using the initial conditions.

We get a = -0.3 m/s and b = 0.1 m/s, which match the values given in the problem.

Hence, The answer is x = 5.6 m

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