A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant 4.0-N force

Question

A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant 4.0-N force for 0.20 m. You then remove your hand, and the cart slides 0.35 m and strikes the 0.50-kg cart. What is the work done by you on the two-cart system? How far does the system’s center of mass move while you are pushing the 1.0-kg cart? By what amount does your force change the kinetic energy of the system’s center of mass?

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Thiên Di 1 month 2021-08-04T09:21:19+00:00 1 Answers 3 views 0

Answers ( )

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    2021-08-04T09:22:32+00:00

    Answer:

    a)W=0.8J

    b) d_t=0.20m

    c) \triangle K.E=0.267J

    Explanation:

    From the question we are told that:

    Mass of cart 1 M_1=1.0kg

    Mass of cart 1 M_2=0.05kg

    Force on  cart 1 F_1=4.0N

    Push Distance of cart 1 d_1=0.20m

    Slide Distance of cart 1 d_1'=0.35m

    a)

    Generally the equation for work-done is mathematically given by

    W=f*d\\W=4*0.20\\W=0.8J \\

    b)

    The systems center of mass moved a net totally of (while being pushed)

    Mass 1 =0.20m

    Mass 2=0

    Therefore

    d_t=d_1+d_2

    d_t=0.20+0

    d_t=0.20m

    c)

    Since work-done is equal to K.E energy of cart 1

    Therefore

    W=1/2mv^2

    V_1=\sqrt{\frac{W}{1/2m}}

    V_1=\sqrt{\frac{0.8}{1/2(1)}}

    V_1=1.264

    Therefore Kinetic energy before collision is

    K.E_1=1/2mv^2

    K.E_1=1/2*1*1.264^2

    K.E_1=0.768

    Generally from the equation for conservation  of momentum the Velocity of cart 2 is mathematically given by

    v_2=\frac{m_1V_1}{m_1+m_2}

    v_2=\frac{1*1.264}{1+0.5}

    V_2=0.842m/s

    Therefore the final K.E is mathematically given by

    K.E_2=(1/2)(m_1+m_2)V_2^2

    K.E_2=1/2*(1.5)(0.842)^2

    K.E_2=0.531J

    Generally the Change in K.E is mathematically given by

    \triangle K.E=K.E_1-K.E_2

    \triangle K.E=0.798-0.531

    \triangle K.E=0.267J

    Therefore the will force change the kinetic energy of the system’s center of mass by

    \triangle K.E=0.267J

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