A box of books is initially at rest a distance D = 0.463 m from the end of a woo

Question

A box of books is initially at rest a distance D = 0.463 m from the end of a wooden board. The coefficient of static friction between the box and the board is ?s = 0.353, and the coefficient of kinetic friction is ?k = 0.288. The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board.

Explanation / Answer

? = angle of elevation of the board. The frictional forces are: ƒs = (Us)?•?(F?) and ƒ? = (U?)?•?(F?) ƒs = (0.301)?•?(?m?•?g?•?cos[?]?) ƒ? = (0.226)?•?(?m?•?g?•?cos[?]?) Fi = initial force in the direction of motion > ƒs Fi = ?m?•?g?•?sin[?] > ƒs ??m?•?g?•?sin[?] > (0.301)?•?(?m?•?g?•?cos[?]?) tan[?] > 0.301 ? > 16.75º ... to start motion Once motion begins the net force on the box (F) in the direction of motion is: F = Fi - ƒ? F = m?•?g?•?sin[?] - (0.226)?•?(?m?•?g?•?cos[?]?) F = m?•?g?•?sin[16.75] - (0.226)?•?(?m?•?g?•?cos[16.75]?) Since F = m?•?a then a = F / m ... in the direction of motion ... dividing by "m" ... a = g?•?sin[16.75] - (0.226)?•?(?g?•?cos[16.75]?) ... next the old physics equation ... V² - (Vi)² = 2?•?a?•?d ... initial velocity = Vi = 0 ????V² - (0)² = 2?•?{ g?•?sin[16.75] - (0.226)?•?(?g?•?cos[16.75]?) }?•?d ?????V² = 2?•?{ (9.8)?•?sin[16.75] - (0.226)?•?(?9.8)?•?cos[16.75]? }?•?(0.743) ?????V² = 2?•?{ 2.825 - 2.121? }?•?(0.743) ???V = 1.02 m/sec

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