Question

A \(5\,\text{kg}\) block pushed by \(100\,\text{N}\) over \(10\,\text{m}\) on a plane. If the coefficient of friction between the block and plane is \(0.2\), then the final kinetic energy of the block is _ _ _ (\(g=10\,\text{m s}^{-2}\)):

• A. \(1000\,J\)
• B. \(900\,J\)
• C. \(800\,J\)
• D. \(700\,J\)

Hint:

Always remember: \[ \text{Net Work} = \text{Change in Kinetic Energy} \] and friction always does negative work because it opposes motion.

Answer 1

The Correct Option is B

Step 1: Use the work energy idea.
The final kinetic energy of a block that starts from rest equals the total work done on it. The push adds energy, but friction takes some away. So we find both and subtract.

Step 2: Work done by the push.
The applied force is $100$ N over a distance of $10$ m. \[ W_{push} = F s = 100 \times 10 = 1000\,\text{J} \]
Step 3: Find the friction force.
On a flat surface the normal force equals the weight. \[ N = mg = 5 \times 10 = 50\,\text{N} \] So friction is \[ f = \mu N = 0.2 \times 50 = 10\,\text{N} \]
Step 4: Work taken by friction.
Friction acts over the same $10$ m but against the motion. \[ W_{friction} = f s = 10 \times 10 = 100\,\text{J} \]
Step 5: Net work on the block.
\[ W_{net} = 1000 - 100 = 900\,\text{J} \]
Step 6: State the kinetic energy.
Since it started from rest, all this net work becomes kinetic energy. \[ \boxed{900\,\text{J}} \]