A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target h

Question

A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.50 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?

in progress 0
Dulcie 7 months 2021-07-17T16:27:37+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-17T16:28:46+00:00

    Answer:

    K = 0.076 J

    Explanation:

    The height of the target, h = 0.860  m

    The mass of the steel ball, m = 0.0120 kg

    Distance moved, d = 1.50 m

    We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

    h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}

    Put all the values,

    t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s

    The velocity of the ball is :

    v=\dfrac{1.5}{0.418}\\\\=	$$3.58\ m/s

    The kinetic energy of the ball is :

    K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J

    So, the required kinetic energy is 0.076 J.

Leave an answer

Browse

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