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,

$a=\frac{v^2-u^2}{2d}$

On substituting the known values,

$a=\frac{0-5^2}{2\times0.8}=-15.62m/s^2$

The frictional force acting on it can be calculated as

$F=ma$

Substituting the known values,

$F=2.0\times-15.62=-31.24\text{ N}$

The energy lost due to friction can be calculated as,

$E=F\times d$

Therefore the energy lost due to friction is,

$E=-31.24\times0.80=-24.99\approx-25\text{ J}$

The negative sign indicates th