a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of the spring

Question

a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of the spring from its unstrained length is -0.115 m, what is the acceleration a of the object

in progress 0
Ngọc Khuê 2 days 2021-07-22T21:30:30+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-07-22T21:31:59+00:00

    Answer:

    The acceleration of the object is 5.57\ m/s^2.

    Explanation:

    Given that,

    Mass of the block, m = 2.6 kg

    Spring constant of the spring, k = 126 N/m

    At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

    When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

    kx=ma

    a is acceleration of the object

    a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2

    So, the acceleration of the object is 5.57\ m/s^2.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )