A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (point A). Whe

Question

A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.55 m up the incline from A, the block is moving up the incline at a speed of 7.50 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is \mu_k = 0.50. The mass of the spring is negligible.
Calculate the amount of potential energy that was initially stored in the spring. Take free fall acceleration to be 9.80 m/s².

in progress 0
Cherry 5 months 2021-08-24T22:13:03+00:00 1 Answers 33 views 0

Answers ( )

    0
    2021-08-24T22:14:41+00:00

    Answer:

    The amount of potential energy that was initially stored in the spring is 88.8 J.

    Explanation:

    Given that,

    Mass of block = 1.60 kg

    Angle = 30.0°

    Distance = 6.55 m

    Speed = 7.50 m/s

    Coefficient of kinetic friction = 0.50

    We need to calculate the amount of potential energy

    Using formula of conservation of energy between point A and B

    U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

    U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

    U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2

    Put the value into the formula

    U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

    U_{A}=88.8\ J

    Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

Leave an answer

Browse

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