Question

A 2.0-kg block is sliding up a rough surface with an initial speed of 5.0 m/s. The block moves up a vertical distance of 0.80 meters before coming to a stop. What is the change in thermal energy of the block? In other words, how much energy did friction take out of the system?

Answer 1

Given,

Mass of the block, m=2.0 kg

The initial speed of the block, u=5.0 m/s

The distance traveled by the block, d=0.80 m/s

As the block comes to rest at the end of the motion, the final velocity of the block, v=0 m/s

From the equation of the motion,

`v^2-u^2=2ad`

Where a is the acceleration of the block.

Rearranging the above equation,

`a=(v^2-u^2)/(2d)`

On substituting the known values,

`a=(0-5^2)/(2*0.8)=-15.62m/s^2`

The frictional force acting on it can be calculated as

`F=ma`

Substituting the known values,

`F=2.0*-15.62=-31.24\text{ N}`

The energy lost due to friction can be calculated as,

`E=F* d`

Therefore the energy lost due to friction is,

`E=-31.24*0.80=-24.99\approx-25\text{ J}`

The negative sign indicates th