bob pushes a crate of mass, m on an incline making an angle theta relative to th

Question

bob pushes a crate of mass, m on an incline making an angle theta relative to the ground. Starting at rest, he pushes with a force, F direct horizontally on the crate. What is the speed the crate when it reaches a height, h above the ground? Ignore Friction.Find the displacement vector in component form as a function of h and theta. Find the vector form of the pushing force, F and calculate the work done by this force. Find the vector form of the gravitational force, Fg and calculate the work done by this force. What is the work done by the normal force of the incline on the crate? why do you not need to find the vector form of this foce? find the total work done on the crate and the speed of the crate. (the horizontal pushing force is in respect to the incline, it is horizontal to the incline)

Explanation / Answer

let horizontal be x axis and vertical be y axis.

let i be the unit vector along x axis and j be the unit vector along y axis.

let origin be the point where the incline touches the ground.

then when the crate rises to a distance of h, then its position is given by

h*cos(theta) i + h*sin(theta) j

pushing force is acting along horizontal direction. i.e. x axis.

so pushing force is given by F i.

work done =dot product of force and displacement=F*h*cos(theta)

vector form of gravitation force, which is acting along vertically downward direction=-m*g j

work done by this force=dot product of force and dispalcement

=-m*g*h*sin(theta)

as normal force is perpendicular to the displacement , dot product of force and displacement is 0.

hence work done by normal force is 0.

total work done on the crate=work done by the force F + work done by the weight of the crate

=F*h*cos(theta)-m*g*h*sin(theta)

using energy conservation principle,

total initial energy + work done =total final energy

as the crate was at height 0 and at rest, total potential energy and kinetic energy were 0.

hence total initial energy=0

work done=F*h*cos(theta)-m*g*h*sin(theta)

if speed after distance h is v,

then 0.5*m*v^2=F*h*cos(theta)-m*g*h*sin(theta)

==>v=sqrt(2*(F*h*cos(theta)-m*g*h*sin(theta))/m)

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