Question

A weight of 500N is pushed up a platform included at 30% to the horizontal. It the efficiency of the plane is 60%, the effort needed to push the weight up the platform is
A. 300N
B. 417N
C. 833N
D.1677N

Answer 1

300newton

I think so

Answer 2

Let AC be the length of inclined plane = 10m.

BC be the height through which the load is lifted = 1m.

Load (L) = 500N.

Effort (E) = L•sin(theta) = 500×1/10= 50N

Therefore,

Mechanical Advantage (MA) = L/Lsin(theta)

MA = 500/50 = 10.

Now, Velocity Ratio (VR) = l/h = 10/1 = 10.

Efficiency = MA/VR=1 (or 100%).

Considering real life scenario.

But, E = 100N (from question)

This is due to Friction between the surface of inclined plane and Load.

Thus, we apply extra 50N.

Now,

MA = 500/100 = 5.

VR remains unchanged.

Hence,

Efficiency = MA/VR = 5/10 = 0.5 (or 50%).

Hope you like it !