Question

A body of mass 10 kg slides down an inclined plane from rest. Its height from the ground level is 10 m. The inclined plane is not smooth. When the body reaches the ground its speed is 14 m/s. Then how much work is done against friction ? $(g=10m/s^2)$

• A. $980J$
• B. $1000J$
• C. $20J$
  
• D. $100J$

Answer 1

The correct option is C $20J$


At point A block will have P.E and as it slides down the inclined plane P.E is converted to K.E

P.E at point A $=mgh=10\times 10\times \times 10=1000J$
K.E at point B $=\frac{1}{2}m{v}^2=\frac{1}{2}\times 10\times \left(14\right)\left(14\right)=5\times 196=980J$

Work Done against friction $1000-980=20J$