Anonymous asked in Science & MathematicsPhysics · 1 decade ago

A 20kg block is dragged over a rough horizontal surface by a constant force of 107N acting at an agle of 25.9?

degrees above the horizontal. The block is displaced 97.8 m, and the coefficient of kinetic friction is 0.142. The acceleration of gravity is 9.8m/s^2. find the magnitude of the work done by the force of friction. Answer in hte units of J.

2 Answers

  • 1 decade ago
    Favorite Answer

    Sketch a free-body diagram for the mass. Coordinate system has +x perpendicular to the surface in the direction of motion; +y normal to the surface.

    The normal force N, on the +y axis

    The friction force μN, on the -x axis

    Gravity mg, on the -y axis

    The pulling force, with components

    P cos 25.9 on the +x axis

    P sin 25.9 on the +y axis

    There's no net force on the vertical axis, so

    N + P sin 25.9 - mg = 0

    Solve for N.

    On the horizontal axis,

    F = ma = P cos 25.9 - μ N

    Solve for F. Then the work W = F d. This work is done to overcome friction, so it's also equal to the work done by friction.

  • Anonymous
    1 decade ago

    1) Draw your Free Body Diagram. The forces on your block are the 107N force F, friction, the normal force, and the gravitational force (or weight).

    2) Write your equation for work: W=f*d*cos(theta), where f is the friction force, d is displacement, and theta is the angle between force and displacement.

    3) Now you know what quantities you have and which you don't. You need to calculate friction, but you already have displacement and theta.

    3a) Write the equation for friction force: f=N*mu, where N is the normal force, and mu is the frictional coefficient. You know you have mu, but need to find N.

    3aa)Write Newton's 2nd law on the y-axis, F=ma, F being the resultant force on the y-axis. If you ask 'why the y-axis' it's because it's the only axis with only ONE unknown (the x-axis gives you two unknowns, but you can only find one at a time). The vertical component of F is F*sin(theta); you also have the vertical Normal Force N, and the block weight m*g: F*sin(theta)+N-m*g=0 (you have zero acceleration in the y-axis because the block is not falling, but is held on the horizontal surface). This gives you N=m*g-F*sin(theta), all quantities you already know so it's plug-and-chug.

    Finally, plug N into the friction force above. Plug Friction into Work and you've got your result.

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