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

Concept: According to the Work-Energy Theorem: \[ W_{\text{net}}=\Delta K \] where \(W_{\text{net}}\) is the net work done on the body and \(\Delta K\) is the change in kinetic energy. The total work done is equal to: \[ W_{\text{applied}}-W_{\text{friction}} \]

Step 1: Calculate the work done by the applied force. Applied force: \[ F=100\,N \] Displacement: \[ s=10\,m \] Hence, \[ W_{\text{applied}}=Fs \] \[ =100\times10 \] \[ =1000\,J \]

Step 2: Calculate frictional force. Coefficient of friction: \[ \mu=0.2 \] Normal reaction: \[ N=mg=5\times10=50\,N \] Thus friction: \[ f=\mu N \] \[ =0.2\times50 \] \[ =10\,N \]

Step 3: Calculate work done against friction. \[ W_{\text{friction}}=fs \] \[ =10\times10 \] \[ =100\,J \]

Step 4: Apply Work-Energy theorem. Net work: \[ W_{\text{net}}=1000-100 \] \[ =900\,J \] Therefore final kinetic energy gained: \[ \boxed{900\,J} \]