A 230kg crate hangs from the end of a rope of length L =12.0M. You push horizont
ID: 2142916 • Letter: A
Question
A 230kg crate hangs from the end of a rope of length L =12.0M. You push horizontally on the crate with force F and the rope swings reaching maximum angle of thata=19.5. from the vertical. Find:-a) The work done by the gravitational force for this displacement.
b) The net work done on the crate to move it from its inital position maximum angular displacement.
A 230kg crate hangs from the end of a rope of length L =12.0M. You push horizontally on the crate with force F and the rope swings reaching maximum angle of thata=19.5. from the vertical. Find:-
a) The work done by the gravitational force for this displacement.
b) The net work done on the crate to move it from its inital position maximum angular displacement.
Explanation / Answer
a) Work done by gravitation=F_grav*deltah=-m*g*deltah=-m*g*(L-L*cos(theta))=-230*9.81*(12-12*cos(19.5/180*pi))=-1553 Joules
Note, here we used F_grav=-m*g with the negative sign because the weight force is in the negative direction (down), and our position axis is directed upwards, thus the vetical displacement is positive and equal to deltah=L-L*cos(theta).
b) Conservation of energy requires:
Work done by the pusher=Final Potential Energy - Initial Potential Energy=m*g*deltah=m*g*(L-L*cos(theta))=1553 Joules
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