Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

Force and Friction Question?

A 15kg box is pulled along a rough surface with force 35N at a 40 degree angle. What is the horizontal and vertical components of this applied force. What is the force of gravity on the box? Normal Force? If the coefficient of the kinetic friction is .25, what is the net force and acceleration? 

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  • 1 month ago
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    Assuming that the 35N force is at an angle of 40° above the horizontal we have:

    (→) 35cos40°

    (↑) 35sin40°

    The force due to gravity is 15g Newtons

    As there is no motion vertically we can say:

    N + 35sin40° = 15g...where N is the normal force

    so, N = 15g - 35sin40° => 124.5 Newtons

    Considering horizontal motion and using F = ma we have:

    35cos40° - Friction = 15a

    Now, Friction = µN => (0.25)(124.5)

    Hence, net force is 35cos40° - (0.25)(124.5) = -4.3 N

    As the net force is negative, the friction force is too great to allow motion to occur.

    Note: 35cos40° > friction for motion to occur

    :)>

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