A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.50 m, 5.30 m) w

Question

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.50 m, 5.30 m) while a constant force acts on it. The force has magnitude 7.40 N and is directed at a counterclockwise angle of 104.° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

in progress 0
Doris 5 years 2021-08-17T03:42:14+00:00 1 Answers 16 views 0

Answers ( )

  1. Thunguyet
    0
    2021-08-17T03:43:14+00:00
    Reply

    Answer:

    The work done by the force is 35.37 Joules.

    Explanation:

    Given that,

    coin started sliding from origin (0,0) to point (1.50m,5.30m)

    therefore displacement of coin along x-axis,x = 1.50m

    displacement of coin along y-axis,y = 5.30m

    angle between applied force and positive x-axis = 104

    applied force = 7.40N

    Now we will find

    component of force along x-axis

    F_{x} = 7.4Ncos104F_{x}=-1.7902

    component of force in y direction,

    F_{y} = 7.4sin 104F_{y}=7.1082N

    work is done by the force on the coin during the displacement will be given as

    W=F_{x}\times x+F_{y}\times yW = -1.7902\times 1.5m + 7.1802N\times 5.3W=35.37Joule

    So, the work done by the force is 35.37 Joules.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )