A 2 kg mass starts from rest on an inclined smooth surface with inclination 30° and length 2 m. How much will it travel before coming to rest on a surface with coefficient of friction 0.25

The correct answer is 4 m.

This problem is solved using the Work-Energy Theorem (or equivalently, conservation of energy). The mass gains kinetic energy while sliding down the smooth incline, and then loses all of it to friction on the horizontal surface.

Step 1: Find the vertical height of the incline.

The incline has a length of 2 m and an inclination of 30°. The vertical height h through which the block descends is:

h=L\times \sin (30°)=2\times \frac{1}{2}=1\text{m}

Step 2: Find the kinetic energy at the bottom of the incline.

Since the inclined surface is smooth (frictionless), all the potential energy converts to kinetic energy. Taking g = 10 m/s²:

KE=mgh=2\times 10\times 1=20\text{J}

Step 3: Find the friction force on the horizontal surface.

On the horizontal surface, the normal reaction equals the weight of the block. The frictional force opposing motion is:

f=\mu mg=0.25\times 2\times 10=5\text{N}

Step 4: Find the distance travelled before coming to rest.

The block comes to rest when all its kinetic energy is used up against friction. If d is the distance travelled on the horizontal surface:

f\times d=KE

5\times d=20

d=\frac{20}{5}=4\text{m}

Conclusion: The block will travel 4 m on the horizontal rough surface before coming to rest.