A 16 kg block is dragged over a rough, horizontal surface by a constant force of 188 N acting at an angle of 33.2 ◦ above the horizontal. Th

Question

A 16 kg block is dragged over a rough, horizontal surface by a constant force of 188 N acting at an angle of 33.2 ◦ above the horizontal. The block is displaced 96.1 m, and the coefficient of kinetic friction is 0.147. Find the magnitude of the work done by the force of friction.

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Tryphena 3 years 2021-08-30T12:09:45+00:00 2 Answers 35 views 0

Answers ( )

  1. bonexptip
    0
    2021-08-30T12:11:04+00:00
    Reply

    Answer:W=-762.84\ J

    Explanation:

    Given

    mass of block m=16\ kg

    Force F=188\ N

    Force is applied at an angle of \theta =33.2^{\circ}

    Displacement of block s=96.1\ m

    coefficient of kinetic friction \mu _k=0.147

    Friction force acting on block f_r

    f_r=\mu _kN

    where N=Normal reaction

    N=mg-F\sin \theta

    N=16\times 9.8-188\times \sin (33.2)

    N=156.8-102.94

    N=53.859\approx 54\ N

    f_r=0.147\times 54

    f_r=7.938\ N

    Work done by friction

    W=f_r\cdot s

    W=7.938\times 96.1\times \cos (180)

    W=-762.84\ J

    Negative sign indicates that displacement is against the direction of force.

  2. Adela
    0
    2021-08-30T12:11:44+00:00
    Reply

    Answer:

    Magnitude of work done W = 2.214 kW.

    Explanation:

    Given :

    Mass of object , m = 16 kg.

    Angle from horizon , \theta=33.2^o.

    Force applied , F = 188 N.

    Displacement of block , D = 96.1 m.

    Coefficient of kinetic friction , \mu=0.147 .

    Component of force, required to move the block, F\ cos\theta=188\times cos\ 33.2^o=157.31\ N.

    Also , frictional force , f=\mu N=\mu (mg)=0.147\times 16\times 9.8=23.045\ N.

    We know, work done , W = Fdcos\theta   ( Here \theta = 180^o , because the direction of displacement is opposite the direction of frictional force .)

    Putting all value in above equation :

    We get , W=23.045\times 96.1\times cos\ 180^o=-2214.62 \ W=-2.214\ kW.

    Magnitude of work done W = 2.214 kW.

    Hence, this is the required solution.

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