A heat engine of efficiency 40% takes 10kJ of heat energy per second. Find the time required by this engine to lift a weight of 1000kg to a height of 10m from the ground. (Take `g= 10ms^(-2)`)

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the power output of the heat engine.
The efficiency (η) of the heat engine is given as 40%, which can be expressed as a decimal:
\[ \eta = 0.40 \]
The heat energy taken by the engine per second (Q) is 10 kJ/s, which is equivalent to:
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