A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass 3 kg that is at rest on a horizontal surf

Question

A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2. After the collision the bullet becomes embedded in the block. How much work is being dne by bullet?

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Minh Khuê 2 months 2021-07-19T07:43:38+00:00 1 Answers 0 views 0

Answers ( )

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    2021-07-19T07:44:41+00:00

    Answer:

    Work done by the bullet is 612.26 J.

    Explanation:

    mass of bullet, m = 0.5 kg

    initial velocity of bullet, u = 50 m/s

    coefficient of friction = 0.2

    mass of block, M = 3 kg

    let the final speed of the bullet block system is v.

    use conservation of momentum

    Momentum of bullet + momentum of block = momentum of bullet block system

    0.5 x 50 + 3 x 0 = (3 + 0.5) v

    v = 7.14 m/s

    let the stopping distance is

    The work done is given by change in kinetic energy of bullet

    initial kinetic energy of bullet, K =  0.5 x 0.5 x 50 x 50 = 625 J

    Final kinetic energy of bullet, K’ = 0.5 x 0.5 x 7.14 x 7.14 = 12.74 J

    So, the work done by the bullet

    W = 625 – 12.74 = 612.26 J  

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