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Math Help - Mechanics forces acting on a object in a slope

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    Mechanics forces acting on a object in a slope

    Mechanics forces acting on a object in a slope-123.jpg answer to q15 is 2.42ms can anyone solve these questions please
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    Re: Mechanics forces acting on a object in a slope

    Quote Originally Posted by abdulrehmanshah View Post
    Click image for larger version. 

Name:	123.JPG 
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ID:	25548 answer to q15 is 2.42ms can anyone solve these questions please
    First, one question per thread.

    Second, these are all Newton's 2nd Law problems. Can you show us how you set this up?

    -Dan
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    Re: Mechanics forces acting on a object in a slope

    well can u please solve question 15 only once u solve it i will make change the topic to solved ok i copied it from a book
    Quote Originally Posted by topsquark View Post
    First, one question per thread.

    Second, these are all Newton's 2nd Law problems. Can you show us how you set this up?

    -Dan
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    Re: Mechanics forces acting on a object in a slope

    forces acting on mass A ...

    T - Mg\sin{\theta} = Ma

    forces acting on mass B ...

    mg - T = ma

    combining equations ...

    mg - Mg\sin{\theta} = Ma + ma

    \frac{m - M\sin{\theta}}{M + m} \cdot g = a

    this acceleration will be constant as mass A slides up 2.8 m ... use an appropriate kinematics equation to find the velocity of mass A when mass B hits the ground.

    for the last 0.2 m, the only force acting on mass A will be the component of weight acting parallel to the incline, so mass A's acceleration will change ...

    g\sin{\theta} = a_2

    use the velocity found when mass B hits the ground, the acceleration a_2, and the remaining displacement to find the final velocity of mass A when it hits the pulley.
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    Re: Mechanics forces acting on a object in a slope

    this is where i got to but i cant get to the final velocity as 2.42ms^-1 can u please help me here
    i write equation 2as=v^2-u^2.
    2*4.9*0.2=v^2-u^2
    what should be u(i.e initial velocity)
    Quote Originally Posted by skeeter View Post
    forces acting on mass A ...

    T - Mg\sin{\theta} = Ma

    forces acting on mass B ...

    mg - T = ma

    combining equations ...

    mg - Mg\sin{\theta} = Ma + ma

    \frac{m - M\sin{\theta}}{M + m} \cdot g = a

    this acceleration will be constant as mass A slides up 2.8 m ... use an appropriate kinematics equation to find the velocity of mass A when mass B hits the ground.

    for the last 0.2 m, the only force acting on mass A will be the component of weight acting parallel to the incline, so mass A's acceleration will change ...

    g\sin{\theta} = a_2

    use the velocity found when mass B hits the ground, the acceleration a_2, and the remaining displacement to find the final velocity of mass A when it hits the pulley.
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    MHF Contributor ebaines's Avatar
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    Re: Mechanics forces acting on a object in a slope

    I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have  \small\Delta KE + \Delta PE = 0, where \small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2 and  \small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m, so you can solve for v_1 . For phase 2:  \small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)
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    Re: Mechanics forces acting on a object in a slope

    what is the value of m
    Quote Originally Posted by ebaines View Post
    I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have  \small\Delta KE + \Delta PE = 0, where \small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2 and  \small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m, so you can solve for v_1 . For phase 2:  \small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)
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    Re: Mechanics forces acting on a object in a slope

    K.E=0.5*m*V where V is final velocity what have u done with phase 2
    Quote Originally Posted by ebaines View Post
    I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have  \small\Delta KE + \Delta PE = 0, where \small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2 and  \small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m, so you can solve for v_1 . For phase 2:  \small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)
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    MHF Contributor ebaines's Avatar
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    Re: Mechanics forces acting on a object in a slope

    "m" stands for "mass." You are told that the mass of block A is 4 Kg and the mass of block B is 3 Kg.

    The change in kinetic energy of an object is:  \frac 1 2 mv_2^2 - \frac 1 2 mv_1^2.
    Last edited by ebaines; November 5th 2012 at 01:19 PM.
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    Re: Mechanics forces acting on a object in a slope

    for the initial motion 2.8 m up the incline ...

    v_f^2 = v_0^2 + 2a(\Delta x)

    v_f = \sqrt{2a(\Delta x)} = \sqrt{\frac{2g(2.8)}{7}} = \, 2.8 m/s

    for the final 0.2 m up the incline ...

    v_f^2 = 2.8^2 - g(0.2)

    v_f = 2.42 \, m/s
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  11. #11
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    Re: Mechanics forces acting on a object in a slope

    how to change the topic to solved
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