Question

A 50 kg block moving initially slideing on a level surface at 20 m/s is subject to a constant friction of 10N untill it comes to a stop. How far will it slide and how much energy is dissipated due to the friction

Answer 1

To know how far the block slides, we need to know it's acceleration in order to know when it stops. You can find the acceleration using Newton's Second Law:

\[F=ma\]\[-10=50a\]\[a=-0.2\]
So the acceleration of the block is -0.2 m/s^2. Remember, friction is acting in the direction opposite the motion (hence -10 in the equation), which means the acceleration is also negative (in fact, it's a deceleration).

We have acceleration [a], initial speed [v(i)], final speed [v(f)] (it's 0 m/s because the block stops), and we want to find distance travelled. That means you can use this equation:

\[v_f^2=v_i^2+2ad\]\[0=400-2(0.2)d\]\[d=1000\]
That means it will take 1000 m for the block to stop moving. The energy dissipated by friction can be calculated either as the work done by friction on the block:

\[W=Fd\]\[W=10(1000)\]\[W=10000\]
Or as the kinetic energy of the block initially (because friction converts all this energy to heat):

\[KE={1 \over 2}mv^2\]\[KE={1 \over 2}(50)(20)^2\]\[KE=10000\]

Either way, you get 10 000 J as the energy dissipated by friction!

If you have any questions about any of this please let me know!

Answer 2

How did you get the 400

Answer 3

Initial velocity squared - initial velocity is 20 m/s in the question, 20 squared is 400.

Answer 4

Oh sorry my bad

Answer 5

No problem!