A hydraulic system is being used to lift a 1500-kg car. If the large piston under the car has a diameter of 50 cm, the small piston has a diameter of 4.0 cm, and the car is lifted a distance of 1.3 m, how much work is done on the car

W = 122.3 J

Explanation:

First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₂ = F₁(A₂/A₁)

where,

F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)

F₁ = 14700 N

F₂ = Force applied to smaller piston = ?

A₁ = Area of larger piston = πd₁²/4

A₂ = Area of smaller piston = πd₂²/4

Therefore,

F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]

F₂ = (14700 N)(d₂²/d₁²)

where,

d₁ = diameter of large piston = 50 cm

d₂ = diameter of small piston = 4 cm

Therefore,

F₂ = (14700 N)[(4 cm)²/(50 cm)²]

F₂ = 94.08 N

Now, for the work done on the car:

Work Done = W = F₂ d

where,

d = displacement of car = 1.3 m

Therefore,

W = (94.08 N)(1.3 m)

W = 122.3 J

v=ir = 3x15 =45 v

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.