physics question

Question

An 8.0kg block is being pulled up a rough incline by a 20 N force. The incline makes an angle of 37 with respect to the horizontal and the coefficient of kinetic friction between the block and the incline is K = 0.3. If the box is initially moving up the incline with a speed of 21m/s, what is the velocity of the block 5 seconds later?

Answer 1

The component of the force of gravity acting down the plan is r.oughly 80N(sin37) = 48N.

The frictional force would be 0.3(80N)cos37=19.6N

That means the net force on the object is down the plan (20N -48N- 19.6N = -47.2N) the acceleration is therefore down the incline -47.2N/8 kg=-5.9m/s².

In 5 seconds the velocity would drop by 29.5 m/s and the block would be moving down the incline at 9.5 m/s. However since it switched directions at the instant itsvelocity is zero, the friction changes direction and is now up the plane and the acceleration is now (20N + 19.6N -48 N)/8 kg = -1.1 m/s².

-20 m/s/ -5,9 m/s² = 3.4 s and there is still 1.6 s left at - 1.1 m/s² and velocity drops (-1,1m/s²*3.4s) 3.7 m/s down the incline.

I used 10 N/kg = 10 m/s² and assumed the 20 N force was up and parallel to the incline for these quickie computations.