Question

1. a forklift is used to lift a large 704 kg crate a distance of 5.54 m in 6.12 s. what is the efficiency of the forklift if it takes a total of 8040 j of chemical energy in the form of burning fuel to complete this task?

Explanation:

Step1: Calculate useful work done (lifting the crate)

The work done in lifting an object is given by \( W = mgh \), where \( m = 704\space kg \), \( g = 9.8\space m/s^2 \), and \( h = 5.54\space m \).
\[ \begin{align*} W_{useful}&=704\times9.8\times5.54\\ &=704\times54.292\\ &= 704\times54.292\\ &= 38221.568\space J \end{align*} \]
Wait, this can't be right as the total energy is 8040 J. Wait, maybe I misread the problem. Wait, the total chemical energy is 8040 J? Wait, no, maybe the height is 5.54 m, mass 704 kg, time 6.12 s, but the total energy input is 8040 J. Wait, no, the useful work should be less than or equal to input energy. Oh, maybe I made a mistake. Wait, let's re - check.

Wait, the formula for work done against gravity is \( W = mgh \). Let's recalculate:

\( m = 704\space kg \), \( g = 9.8\space m/s^2 \), \( h = 5.54\space m \)

\( W_{useful}=704\times9.8\times5.54 \)

First, \( 704\times9.8 = 704\times(10 - 0.2)=7040-140.8 = 6900 - 140.8+140? No, 704\times9.8 = 704\times9 + 704\times0.8=6336+563.2 = 6899.2 \)

Then \( 6899.2\times5.54 \)

\( 6899.2\times5 = 34496 \)

\( 6899.2\times0.54 = 6899.2\times0.5+6899.2\times0.04 = 3449.6+275.968 = 3725.568 \)

Total \( W_{useful}=34496 + 3725.568=38221.568\space J \). But this is more than 8040 J, which is impossible. So there must be a misinterpretation. Wait, maybe the mass is 70.4 kg? Let's check. If \( m = 70.4\space kg \):

\( W_{useful}=70.4\times9.8\times5.54 \)

\( 70.4\times9.8 = 690. - 14.08=675.92 \)

\( 675.92\times5.54 = 675.92\times5+675.92\times0.54 = 3379.6+365.0 = 3744.6\space J \). Still more than 8040? No, 3744 is less than 8040? No, 3744 < 8040? No, 3744 is less than 8040? Wait, 3744 is less than 8040. Wait, maybe the mass is 70.4 kg. Maybe a typo in the problem, 704 kg is too much. Alternatively, maybe the height is 0.554 m. Let's assume that the problem has a typo and the mass is 70.4 kg.

Alternatively, maybe the total energy input is 80400 J? Wait, the original problem: "total of 8040 J of chemical energy". This is a problem. Wait, maybe I misread the mass. Let's check the problem again: "lift a large 704 kg crate a distance of 5.54 m in 6.12 s. What is the efficiency of the forklift if it takes a total of 8040 J of chemical energy".

This is a contradiction because the work done to lift 704 kg by 5.54 m is much more than 8040 J. So there must be an error. Wait, maybe the distance is 0.554 m. Let's try \( h = 0.554\space m \):

\( W_{useful}=704\times9.8\times0.554 \)

\( 704\times9.8 = 6900 - 140.8 = 6759.2 \)

\( 6759.2\times0.554 = 6759.2\times0.5+6759.2\times0.054 = 3379.6+365.0 = 3744.6\space J \). Still, 3744 < 8040? No, 3744 is less than 8040? No, 3744 is less than 8040. Wait, 3744 < 8040. Then efficiency would be \( \eta=\frac{W_{useful}}{W_{total}}\times100\%=\frac{3744}{8040}\times100\%\approx46.57\% \). But this is under the assumption that there is a typo.

Alternatively, maybe the mass is 70.4 kg. Let's try \( m = 70.4\space kg \), \( h = 5.54\space m \):

\( W_{useful}=70.4\times9.8\times5.54 = 70.4\times54.292 = 70.4\times50+70.4\times4.292 = 3520+302.1568 = 3822.1568\space J \)

Then efficiency \( \eta=\frac{3822.16}{8040}\times100\%\approx47.54\% \)

But the problem as stated has a contradiction because 704 kg lifted 5.54 m requires more than 8040 J. So perhaps it's 70.4 kg.

Assuming that it's a typo and the mass is 70.4 kg:

Step1: Calculate useful work (\( W_{useful} \))

\( W_{useful}=mgh \), \( m = 70.4\space kg \), \( g = 9.8\space m/s^2 \), \( h = 5.54\space m \)
\[ \begin{align*} W_{useful}&=70.4\times9.8\times5.54\\ &=70.4\times54.292\\ &= 3822.1568\space J \end{align*} \]

Step2: Calculate efficiency (\( \eta \))

Efficiency is given by \( \eta=\frac{W_{useful}}{W_{total}}\times100\% \), where \( W_{total}=8040\space J \)
\[ \begin{align*} \eta&=\frac{3822.16}{8040}\times100\%\\ &\approx0.4754\times100\%\\ &\approx47.5\% \end{align*} \]

But since the original problem has a contradiction (704 kg lifted 5.54 m needs more than 8040 J), maybe the mass is 70.4 kg. Alternatively, maybe the height is 0.554 m. Let's try \( h = 0.554\space m \) with \( m = 704\space kg \):

Step1: Calculate useful work

\( W_{useful}=704\times9.8\times0.554 \)
\[ \begin{align*} W_{useful}&=704\times5.4292\\ &= 3822.1568\space J \end{align*} \]

Step2: Calculate efficiency

\( \eta=\frac{3822.16}{8040}\times100\%\approx47.5\% \)

So, assuming that there is a typo and the height or mass is different, the efficiency is approximately 47.5% (or depending on the correct values). But since the problem as stated has inconsistent values, but following the given numbers (even though physically impossible), if we proceed with \( m = 704\space kg \), \( h = 5.54\space m \), \( W_{total}=8040\space J \), the useful work[LLM SSE On Failure]

Answer:

Step1: Calculate useful work done (lifting the crate)

The work done in lifting an object is given by \( W = mgh \), where \( m = 704\space kg \), \( g = 9.8\space m/s^2 \), and \( h = 5.54\space m \).
\[

$$\begin{align*} W_{useful}&=704\times9.8\times5.54\\ &=704\times54.292\\ &= 704\times54.292\\ &= 38221.568\space J \end{align*}$$

\]
Wait, this can't be right as the total energy is 8040 J. Wait, maybe I misread the problem. Wait, the total chemical energy is 8040 J? Wait, no, maybe the height is 5.54 m, mass 704 kg, time 6.12 s, but the total energy input is 8040 J. Wait, no, the useful work should be less than or equal to input energy. Oh, maybe I made a mistake. Wait, let's re - check.

Wait, the formula for work done against gravity is \( W = mgh \). Let's recalculate:

\( m = 704\space kg \), \( g = 9.8\space m/s^2 \), \( h = 5.54\space m \)

\( W_{useful}=704\times9.8\times5.54 \)

First, \( 704\times9.8 = 704\times(10 - 0.2)=7040-140.8 = 6900 - 140.8+140? No, 704\times9.8 = 704\times9 + 704\times0.8=6336+563.2 = 6899.2 \)

Then \( 6899.2\times5.54 \)

\( 6899.2\times5 = 34496 \)

\( 6899.2\times0.54 = 6899.2\times0.5+6899.2\times0.04 = 3449.6+275.968 = 3725.568 \)

Total \( W_{useful}=34496 + 3725.568=38221.568\space J \). But this is more than 8040 J, which is impossible. So there must be a misinterpretation. Wait, maybe the mass is 70.4 kg? Let's check. If \( m = 70.4\space kg \):

\( W_{useful}=70.4\times9.8\times5.54 \)

\( 70.4\times9.8 = 690. - 14.08=675.92 \)

\( 675.92\times5.54 = 675.92\times5+675.92\times0.54 = 3379.6+365.0 = 3744.6\space J \). Still more than 8040? No, 3744 is less than 8040? No, 3744 < 8040? No, 3744 is less than 8040? Wait, 3744 is less than 8040. Wait, maybe the mass is 70.4 kg. Maybe a typo in the problem, 704 kg is too much. Alternatively, maybe the height is 0.554 m. Let's assume that the problem has a typo and the mass is 70.4 kg.

Alternatively, maybe the total energy input is 80400 J? Wait, the original problem: "total of 8040 J of chemical energy". This is a problem. Wait, maybe I misread the mass. Let's check the problem again: "lift a large 704 kg crate a distance of 5.54 m in 6.12 s. What is the efficiency of the forklift if it takes a total of 8040 J of chemical energy".

This is a contradiction because the work done to lift 704 kg by 5.54 m is much more than 8040 J. So there must be an error. Wait, maybe the distance is 0.554 m. Let's try \( h = 0.554\space m \):

\( W_{useful}=704\times9.8\times0.554 \)

\( 704\times9.8 = 6900 - 140.8 = 6759.2 \)

\( 6759.2\times0.554 = 6759.2\times0.5+6759.2\times0.054 = 3379.6+365.0 = 3744.6\space J \). Still, 3744 < 8040? No, 3744 is less than 8040? No, 3744 is less than 8040. Wait, 3744 < 8040. Then efficiency would be \( \eta=\frac{W_{useful}}{W_{total}}\times100\%=\frac{3744}{8040}\times100\%\approx46.57\% \). But this is under the assumption that there is a typo.

Alternatively, maybe the mass is 70.4 kg. Let's try \( m = 70.4\space kg \), \( h = 5.54\space m \):

\( W_{useful}=70.4\times9.8\times5.54 = 70.4\times54.292 = 70.4\times50+70.4\times4.292 = 3520+302.1568 = 3822.1568\space J \)

Then efficiency \( \eta=\frac{3822.16}{8040}\times100\%\approx47.54\% \)

But the problem as stated has a contradiction because 704 kg lifted 5.54 m requires more than 8040 J. So perhaps it's 70.4 kg.

Assuming that it's a typo and the mass is 70.4 kg:

Step1: Calculate useful work (\( W_{useful} \))

\( W_{useful}=mgh \), \( m = 70.4\space kg \), \( g = 9.8\space m/s^2 \), \( h = 5.54\space m \)
\[

$$\begin{align*} W_{useful}&=70.4\times9.8\times5.54\\ &=70.4\times54.292\\ &= 3822.1568\space J \end{align*}$$

\]

Step2: Calculate efficiency (\( \eta \))

Efficiency is given by \( \eta=\frac{W_{useful}}{W_{total}}\times100\% \), where \( W_{total}=8040\space J \)
\[

$$\begin{align*} \eta&=\frac{3822.16}{8040}\times100\%\\ &\approx0.4754\times100\%\\ &\approx47.5\% \end{align*}$$

\]

But since the original problem has a contradiction (704 kg lifted 5.54 m needs more than 8040 J), maybe the mass is 70.4 kg. Alternatively, maybe the height is 0.554 m. Let's try \( h = 0.554\space m \) with \( m = 704\space kg \):

Step1: Calculate useful work

\( W_{useful}=704\times9.8\times0.554 \)
\[

$$\begin{align*} W_{useful}&=704\times5.4292\\ &= 3822.1568\space J \end{align*}$$

\]

Step2: Calculate efficiency

\( \eta=\frac{3822.16}{8040}\times100\%\approx47.5\% \)

So, assuming that there is a typo and the height or mass is different, the efficiency is approximately 47.5% (or depending on the correct values). But since the problem as stated has inconsistent values, but following the given numbers (even though physically impossible), if we proceed with \( m = 704\space kg \), \( h = 5.54\space m \), \( W_{total}=8040\space J \), the useful work[LLM SSE On Failure]