Question

A creat of mass 350kg is pulled of a rough inclined plane inclined at an angle of 60degree to the horizontal.determined the total work done in moving the crate to the top of the inclined 20m high given. 1.mass of crate =50kg. 2.weight of crate =500n. 3lenght of inclined plane =l=h_sin =2o_sin60=23. Coefficient of friction mill =tan =tan60degree =1.732. Reaction force to weight ,r=w cos60. Find the total work done. Formula=[wsm"tita" +mill]*d

Answer 1

You say the mass is 350 kg, then you say it is 50 kg ?
I will assume mass = m
You are evidently assuming g = 10 m/s^2. However to be safe I will just call it g.

Component of weight down slope = m g sin 60 = 0.866 m g
Component of weight normal to surface = m g cos 60 = 0.500 m g
friction force down slope mu m g cos 60 = 0.500 m g mu
These are balanced by pull force F up slope
F = (0.866 + 0.500 mu) m g
work done to pull up = F * d = F * 20/sin 60 = 23 F
= 23 (0.866 + 0.500 mu) m g Joules