(a) At a certain instant, a particle-like object is acted on by a force F= (5.7 N)i -(2.7 N)j + (5.0 N)k while the object’s velocity is v=

Question

(a) At a certain instant, a particle-like object is acted on by a force F= (5.7 N)i -(2.7 N)j + (5.0 N)k while the object’s velocity is v= -(2.3 m/s)i + (5.8 m/s)k. What is the instantaneous rate at which the force does work on the object?
(b) At some other time, the velocity consists of only a y component. If the force is unchanged, and the instantaneous power is -8.70 W, what is the velocity of the object just then? (Give your answer without a unit vector.)

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Orla Orla 2 months 2021-07-22T16:55:33+00:00 1 Answers 4 views 0

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    2021-07-22T16:56:37+00:00

    Answer:

    (a) The force which works on the object at a rate 15.89 w.

    (b) The velocity of the object when the power is -8.7 w is (3.22\ m/s)\hat j.

    Explanation:

    Work: The  work on an object is the dot product of force that acts on the object and velocity of the object .

    P = F.V

    Dot product:

    \vec a=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}

    \vec b=b_x\hat{i}+b_y\hat{j}+b_z\hat{k}

    \vec a.\vec b=(a_x\hat{i}+a_y\hat{j}+a_z\hat{k}).(b_x\hat{i}+b_y\hat{j}+b_z\hat{k})

         =a_x.b_x+a_y.b_y+a_z.b_z

    \vec a.\vec b=|a||b|cos\theta

    where the angle between a and bis θ  

    (a)

    Given that,

    \vec F=(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k

    and

    \vec V=-(2.3 m/s) \hat i+(5.8 m/s)\hat j

    \Rightarrow \vec V=-(2.3 m/s) \hat i+(0m/s)\hat j+(5.8 m/s)\hat j

    The work on the object is

    =\vec F.\vec V

    =\{(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k\}.\{-(2.3 m/s) \hat i+(0m/s)\hat j+(5.8 m/s)\hat j\}

    ={(5.7)×(-2.3)+(-2.7)×0+(5.0×5.8)} w

    =15.89 w

    The instantaneous rate at which the force does work on the object is 15.89 w.

    (b)

    The velocity of the object consists of only a y component i.e the x component and y component are zero.

    V_x=0 and V_z=0

    Let,

    \vec V= 0\hat i+ a \hat j+0 \hat k

    The power is -8.70 w.

    \vec F=(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k

    ∴ -8.70 = \{(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k\}.( 0\hat i+ a \hat j+0 \hat k)

    ⇒ – 8.70 = -2.7×a

    \Rightarrow a=\frac{-8.7}{-2.7}

    ⇒ a = 3.22

    The velocity of the object is (3.22\ m/s)\hat j

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