Question

PHYS 1441 Spring 2019 Assignment (Part 2) Problem C42-07 The crate shown in (Figure 1) lies on a plane tited at an angle 6-230 to the horizontal, with 0.19 Units Figure 1 of1 Part B the crane starts tom rest 825 ma up alog te pianefrom is base, what w·bethe crate's speed shor, treaches the boton ofthe i do? Express your answer to two significant figures and include the appropriate units 12 Units

Answer 1

the weight of the crate is acting vertically downwards and the value will be,

W mg

the normal reaction of the box with the plane will be

$N = mgcos(\theta) = mgcos(23)$

so the frictional force will be,

$f = \mu N$

f = 0.19 × m.q.cos(23)

A)

the acceleration component along the plane will be

acceleration- gsinθ-g.su ( )

but the opposing frictional force will reduce this to

$a = g.sin(23) - 0.19\times g.cos(23)$

$\mathbf{a =2.117 m/s^2}$

B)

using eqn of motion, ,

and u = 0 initially

and s = 8.25m

we have

$v^2 = 2\times 2.117\times 8.25$

$v = \sqrt{34.9305}$

$\mathbf{v = 5.91m/s}$