Question

A hydraulic lift is used to jack a 970 kg car 12cm off the floor . The diameter of the output piston is 18cm and the input force is 250N. (a) what is the area of the input piston? (b) what is the work done in lifting the car 12cm? (c) If the input piston moves 13cm in each stroke how high does the car move up for each stroke? (d) How many strokes are required to jack the car up 12cm? (e) show that energy is conserved
So far for (a) My partner and I got .005m and for (b) we did 970x9.8x12= 1140.72J
I don't feel like we are on the right track. Please help! I don't know how to answer the rest

Answer 1

From Pascal's Law, we know that the pressure on the input and outside will be the same. From the definition of pressure, we know that\[P = {F \over A}\]We can relate the input and output force and area as\[{F_i \over A_i} = {F_o \over A_o}\]\[{250 \over A_i} = {(970 \cdot g) \over \left( {\pi \over 4} 18^2 \right) }\]
This is for part a.

For part b. Remember that\[W = F \cdot d = 970 \cdot 9.81 \cdot 0.12\]

For part c. We need to relate the volume displaced by the 13cm input stroke to the volume displaced by the output stroke. Realize that both volumes must be same. Therefore, we get\[A_i s_i = A_o s_o\]where s is the stroke of the input and output sides.

For part d. With knowledge of the output stroke lenght, we can divide that 12cm height that the car is lifted by this stroke to get number of input strokes. This is expressed as\[N = {12 \over s_o}\]

For part e. With knowledge of the number of input strokes, we can use the work-energy theorem to show that energy is conserved. The work energy theorem states thats that work done on a system equals the change in energy of the system. This is expressed as\[W = \Delta E = \Delta KE + \Delta PE\]Therefore, we can show that energy will be conserved when the input work equals the output work\[W_i = W_o\]We already solved for the output work. We can solve for the iutput work as \[W_i = F_i \cdot N \cdot s_i\]