Object speed at the end

Question

The figure shows the height of the hill h=2.8 m. Distance BC length s=2.8 m. A piece with mass m=3 kg is discharged from rest to motion from point A. to point B and the , the speed between AB of the piece is v=4.7 m/s. And the coefficient of friction μ = 0,30. So I should calculate the speed of the piece at point C. But I don't know how...

Answer 1

Ok, we will have a deal: I am showing you the solution of this problem - and you will reveal me the secret: HOW DID YOU MANAGE TO INSERT THE PICTURE INTO YOUR PROBLEM???

When the body is on horizontal part of the surface, the unbalanced force acting on the body - is the kinetic frictional force that causes deceleration of the body and its speed decreasing.

Hence, we can use the formula from kinematic

v ² - v₀² = - 2as

I took "minus" in the formula , because it is decelerated motion and the vector of acceleration of the body has the direction opposite to the direction of motion.

From this formula

v = √ (v₀² - 2as)

To find a , I will use the Newton second law:

f = ma

But we know that frictional force

f = µN

The body is on the horizontal part of the surface and the unbalanced force acts in horizontal direction, hence normal reaction force

N = mg.

Now it is easy to see that

a = µg

So finally, the speed of the body at point C

v = √ (v₀² - 2µgs)

Let's put into this formula all known values and we will get

v = √ ((4.7 m/s)² - 2·0.30·9.8 m/s²·2.8 m)

or

v = 2.4 m/s

Please, in the comments tell me how did you put the picture; it will help me in the future to do the same. These will help many students, who are asking us to show solutions to physics problems. Thanks in advance.