Work Done by Constant and Variable Forces: Practice Problem & Solution
$300\text{ J}$ of work is done in sliding a $2\text{ kg}$ block up an inclined plane of height $10\text{ m}$. Taking $g = 10\text{ m/s}^2$, work done against friction is: (2006)
Solution Explained:
To solve this problem, we apply the core principles of Work Done by Constant and Variable Forces. Understanding the underlying formula is key to arriving at the correct answer below:
Total work done on the block goes into increasing potential energy and overcoming friction. Potential energy gain is $U = mgh = 2 \times 10 \times 10 = 200\text{ J}$. Work done against friction = Total work - $U = 300 - 200 = 100\text{ J}$.
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