Question

a hydraulic system at the mechanic shop can perform up to 180,000 joules of work in a day. if throughout the day it can lift 10 cars from the ground to a height of 2 meters, how much can the cars each weigh in newtons?

Explanation:

Step1: Recall the work formula

The formula for work done against gravity is \( W = F \times d \), where \( W \) is work, \( F \) is force (weight in this case), and \( d \) is distance. Here, total work \( W_{total}= 180000 \, \text{J} \), number of cars \( n = 10 \), height \( d = 2 \, \text{m} \). Let the weight of each car be \( F \).

Step2: Calculate total force for all cars

First, find the total force \( F_{total} \) needed to lift 10 cars. From \( W = F \times d \), we can rearrange to \( F=\frac{W}{d} \). For total work and total distance (total height lifted for all cars: \( d_{total}=n\times d = 10\times2 = 20 \, \text{m} \)), so \( F_{total}=\frac{W_{total}}{d_{total}}=\frac{180000}{20}=9000 \, \text{N} \).

Step3: Find weight of each car

Since there are 10 cars, the weight of each car \( F=\frac{F_{total}}{n}=\frac{9000}{10} = 9000 \, \text{N} \)? Wait, no, wait. Wait, actually, the work done for each car is \( W_{per\ car}=\frac{W_{total}}{n}=\frac{180000}{10} = 18000 \, \text{J} \) per car. Then for each car, using \( W = F \times d \), \( F=\frac{W_{per\ car}}{d}=\frac{18000}{2}=9000 \, \text{N} \). Wait, let's correct the steps.

Alternative Step2: Work per car. Total work is 180000 J for 10 cars, so work per car \( W_{car}=\frac{180000}{10}=18000 \, \text{J} \).

Step3: For each car, \( W_{car}=F\times d \), so \( F = \frac{W_{car}}{d} \). Given \( d = 2 \, \text{m} \), \( F=\frac{18000}{2}=9000 \, \text{N} \)? Wait, no, that can't be. Wait, no, let's re - do.

Wait, the correct approach: The work done to lift one car is \( W = F\times h \), where \( h = 2 \, \text{m} \), \( F \) is weight of one car. For 10 cars, total work \( W_{total}=10\times F\times h \). So we can solve for \( F \):

\( 180000=10\times F\times2 \)

\( 180000 = 20F \)

\( F=\frac{180000}{20}=9000 \, \text{N} \). Wait, yes, that's correct. So each car weighs 9000 Newtons? Wait, no, wait 180000 divided by (10*2) is 180000/20 = 9000. Yes.

Wait, let's check again. Total work: 180000 J. Lifting 10 cars, each to 2 m. So the total distance the force is applied is for each car 2 m, 10 cars: 10*2 = 20 m. Work is force times distance, so force is work over distance: 180000 J / 20 m = 9000 N. That force is the total force to lift 10 cars, so each car is 9000 N /10 = 900 N? Wait, oh no! I made a mistake in step 2. Oh right! Total force for 10 cars is \( F_{total} \), and each car has force \( F \), so \( F_{total}=10F \). Then total work \( W = F_{total}\times d \), where \( d = 2 \, \text{m} \) (because each car is lifted 2 m, so the total force is applied over 2 m, not 20 m). Wait, that's the error. Let's correct:

Correct Step2: The work done to lift 10 cars, each lifted 2 m. So for each car, work \( W_{car}=F\times2 \), so total work \( W_{total}=10\times(F\times2)=20F \).

Step3: Then solve for \( F \): \( 180000 = 20F \), so \( F=\frac{180000}{20}=9000 \)? No, wait, no: \( W_{total}=n\times(F\times d) \), where \( n = 10 \), \( d = 2 \), so \( 180000=10\times F\times2 \), so \( 20F = 180000 \), so \( F = 9000 \)? But that seems too heavy. Wait, no, maybe my initial approach was wrong. Wait, work is force times distance. If you lift one car 2 meters, the work per car is \( F\times2 \). For 10 cars, total work is \( 10\times F\times2=20F \). So \( 20F = 180000 \), so \( F = 9000 \) Newtons. But a car weighs about 1000 kg, so weight is \( mg = 1000\times9.8 = 9800 \) N, so 9000 N is reasonable (approximate). Wait, maybe the numbers are chosen for simplicity. So the correct calculation is:

Given \( W = 180000 \, \text{J} \), \( n = 10 \), \( h = 2 \, \text{m} \).

We know that work done to lift an object is \( W = F\times h \), where \( F \) is the weight (force due to gravity). For \( n \) objects, total work \( W = n\times F\times h \).

So rearranging for \( F \): \( F=\frac{W}{n\times h} \)

Plugging in the values: \( F=\frac{180000}{10\times2}=\frac{180000}{20}=9000 \, \text{N} \)? Wait, no, wait 180000 divided by 20 is 9000? Wait 10*2 is 20, 180000/20 is 9000. Yes. So each car weighs 9000 Newtons? But that's about 900 kg (since \( F = mg \), \( m=\frac{F}{g}=\frac{9000}{9.8}\approx918 \, \text{kg} \)), which is a small car, maybe. So the calculation is correct.

Wait, let's re - derive:

Work formula: \( W = F \cdot d \), where \( F \) is force (weight of the car, since we are lifting against gravity), \( d \) is the distance lifted (height).

Total work done: \( W_{total}=180000 \, \text{J} \)

Number of cars: \( N = 10 \)

Height per car: \( d = 2 \, \text{m} \)

Let weight of each car be \( F \). Then total work is the sum of work done on each car. Work done on one car: \( W_{1}=F\times d \). Work done on 10 cars: \( W_{total}=N\times F\times d \)

So, \( 180000 = 10\times F\times2 \)

\( 180000 = 20F \)

Divide both sides by 20: \( F=\frac{180000}{20}=9000 \, \text{N} \)

Yes, that's correct.

Answer:

The weight of each car is \(\boldsymbol{9000}\) newtons.