A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, w

Question

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.A) How much work is done by the gravitational force on thecrate?
B) Determine the increase in internal energy of the crate-inclinesystem owing to friction.
C) How much work is done by the 100N force on the crate?
D) What is the change in kinetic energy of the crate?
E) What is the speed of the crate after being pulled 5.00m?

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Thu Thủy 3 years 2021-08-21T19:11:51+00:00 1 Answers 90 views 0

Answers ( )

  1. kimchi2
    0
    2021-08-21T19:13:26+00:00
    Reply

    Given that,

    Mass = 9.2 kg

    Force = 110 N

    Angle = 20.2°

    Distance = 5.10 m

    Speed = 1.58 m/s

    (A). We need to calculate the work done by the gravitational force

    Using formula of work done

    W_{g}=mgd\sin\theta

    Where, w = work

    m = mass

    g = acceleration due to gravity

    d = distance

    Put the value into the formula

    W_{g}=9.2\times(-9.8)\times5.10\sin20.2

    W_{g}=-158.8\ J

    (B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

    Using formula of potential energy

    \Delta U=-W

    Put the value into the formula

    \Delta U=-(-158.8)\ J

    \Delta U=158.8\ J

    (C). We need to calculate the work done by 100 N force on the crate

    Using formula of work done

    W=F\times d

    Put the value into the formula

    W=100\times5.10

    W=510\ J

    We need to calculate the work done by frictional force

    Using formula of work done

    W=-f\times d

    W=-\mu mg\cos\theta\times d

    Put the value into the formula

    W=-0.4\times9.2\times9.8\cos20.2\times5.10

    W=-172.5\ J

    We need to calculate the change in kinetic energy of the crate

    Using formula for change in kinetic energy

    \Delta k=W_{g}+W_{f}+W_{F}

    Put the value into the formula

    \Delta k=-158.8-172.5+510

    \Delta k=178.7\ J

    (E). We need to calculate the speed of the crate after being pulled 5.00m

    Using formula of change in kinetic energy

    \Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})

    v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2

    Put the value into the formula

    v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58

    v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}

    v_{2}=6.35\ m/s

    Hence, (A). The work done by the gravitational force is -158.8 J.

    (B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

    (C). The work done by 100 N force on the crate is 510 J.

    (D). The change in kinetic energy of the crate is 178.7 J.

    (E). The speed of the crate after being pulled 5.00m is 6.35 m/s

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