Question

A 5kg body is moving on a rough surface (Us = 0.2) laong a straight line at t=0 its velocity is 10m/s from t=0 a constant force F =20N is applied on the body in the same direction opposite to its velocity also from t=0 the plane surface is pulled along the same direction as the body's velocity with a constant velocity 8m/s Determine the body's velocity after t=4s?

Answer 1

Well we have two methods but the best method is to go like this.
Consider yourself standing on the plane, then relative to you the initial velocity of the body is 2 m/s
\[ F_{net} = -20 - \mu N \]
\[ F_{net} = -20 - 0.2 \cdot N \]
\[ a = \frac{ -20 - 0.2 \cdot mg}{m} \]
Now let's find when the velocity of the body becomes zero
\[ v= u + at \]
\[ 0 = 2 -\frac{ 20}{m} - 0.2 g t \]

And hence you need mass of the body, if mass of the body is unknown the problem can't be solved.

Answer 2

sorry the mass is 5kg

Answer 3

@ammubhave

Answer 4

\[ a = \frac{-20 - 0.2 * 5 * 10}{5} \]
\[ a = \frac{-20 - 10}{5} \]
\[ a = -6 \, m/s^2 \]

Now find where v = 0
\[ 0 = 2 - 6t \]
\[ t = \frac{1}{3} sec \]

Now at this point the block will begin to move in reverse direction and hence the applied force and the frictional force will be in opposite direction.
So now the new acceleration will be:
\[ a = \frac{-20 + 10}{5} \]
\[ a = -2 \, m/s^2 \]
Now the velocity after 4 second from t = 0:
\[ v = 0 - 2\cdot (4 - \frac{1}{3}) \]
\[ = -2(\frac{12 - 1}{3}) \]
\[ = -\frac{22}{3} \, m/s \]

I am not sure if the above calculations are correct but the method will be this only.

Answer 5

it should be 2/3

Answer 6

@experimentX

Answer 7

@Yahoo! The -22/3 is the velocity relative to the plane. If you want the velocity relative to the original coordinate system, you have to add back in the 8m/s.\[\frac{-22}{3}+8=\frac{2}{3}\]

Answer 8

Yes u r correct @eseidl