In a hydraulic jack, the pipe below the load has an area of 5 m2, and the pipe through which the input force transmits has an area of 1 m2. calculate the force required. kn of input force is necessary to balance a load of 10,000 n.

0.2kN

Step-by-step explanation:

An hydraulic jack works by the principle that pressure applied to one part in an enclosed liquid is transmitted equally in all parts of the liquid. this is in accordance with Pascal's principle.

Pressure= Force/Area

P=F/A

Pressure due to the load= Pressure due to input force for the system to be in balance.

Let force due to load be F₁, input force be F₂, Area at the load piston be A₁ and the area of the pipe through which the input force transmits be A₂, then,

F₁/A₁=F₂/A₂

10000N/5m²=F₂/1m²

F₂=(1000N×1m²)/5m²

F₂=200N

=0.2kN

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50