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

Question

. 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?

Answer 1

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

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

2). Speed at the top of the slope,

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