Answer to Question #113640 in Mechanics | Relativity for Hetisani Sewela

Question #113640
. a. Starting from rest, a crate of mass m is pushed up a frictionless slope of angle u by a horizontal force of
magnitude F. Use work and energy to find an expression for the crate’s speed v when it is at height h
above the bottom of the slope.
b. Doug uses a 25 N horizontal force to push a 5.0 kg crate up a 2.0 m high, 20o
frictionless slope. What
is the speed of the crate at the top of the slope?
Expert's answer

Explanations & Calculations

  • Apply work & energy relationship along the slope ahead as the object travel from the bottom to the given height.
  • Supplied work turns into the work done against the component of the object's own weight & the energy stores as the kinetic energy.

1). Consider the notations in the sketch

"\\qquad\n\\begin{aligned}\n\\small F\\cos u*x &= \\small mg\\sin u*x + \\frac{1}{2}mv^2\\\\\n\\small F\\cos u*\\frac{h}{\\sin u} &= \\small mg\\cancel{\\sin u}*\\frac{h}{\\cancel{\\sin u}} + \\frac{1}{2}mv^2\\\\\n\\small F*\\frac{h}{\\tan u} &= \\small mgh + \\frac{1}{2}mv^2\\\\\n\\small v &= \\small \\bold{ \\sqrt{\\frac{2}{m}\\Bigg(F*\\frac{h}{\\tan u}- mgh\\Bigg)}}\n\\end{aligned}"

2). Speed at the top of the slope,

"\\qquad\n\\begin{aligned}\n\\small v &=\\small\\sqrt{\\frac{2}{5kg}\\Bigg(25N*\n\\frac{2m}{\\tan 20}-5kg*9.8ms^{-2}*2m\\Bigg)}\\\\\n\\small &= \\small \\bold{3.969ms^{-1}} \n\\end{aligned}"

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