A 2.0 kgkg wood block is launched up a wooden ramp that is inclined at a 25 ∘∘ angle. The block's initial speed is 11 m/sm/s . The coefficient of kinetic friction of wood on wood is μk=0.200μk=0.200. Part A What vertical height does the block reach above its starting point?

The B part of the question is missing and it says;

What speed does it have when it slides back down to its starting point?

A) vertical height the block reaches above its starting point = 3.568m

B) Speed of block when it slides back down to its starting point = 6.323 m/s

Explanation:

A) Since inclined at an angle of 30°,thus;

For kinetic friction, h = mg cosθ

Thu kinetic friction force(Fk) is ;

Fk = μmg cosθ

Let's choose the x-direction to be parallel to the ramp surface.

Resolving all the forces on the block in the x-direction;

ΣFx; μmg cosθ = -ma - mg sinθ

Divide each term by m to get;

μg cosθ = -a - g sinθ

Making a the subject of the formula,

a = -μg cosθ - g sinθ

a = - g(μ cosθ + sinθ)

So a = - 9.81[0.2(cos25°) + sin 25°]

a = - 9.81[ 0.1813 + 0.4226]

a = -5.924 m/s²

Using equations of motion,

v² = u² + 2as

Now v = 0m/s while u= 10m/s and s=Δx since ramp is chosen to be parallel to x-direction

Thus 0² = 10² + 2(-5.924)(Δx)

So making (Δx) the subject ;

11.85(Δx) = 100

So, Δx = 100/11.85 = 8.44m

Thus; vertical height which is the vertical component displacement = (Δx) x sinθ = 8.44Sin(25) = 8.44 x 0.4226 = 3.568m

B) when the block is sliding down, the sign of the friction changes.

Thus;

ΣFx; - μmg cosθ = -ma - mg sinθ

So, a = μg cosθ - g sinθ

a = g(μ cosθ - sinθ)

So a = 9.81[0.2(cos25°) - sin 25°]

a = 9.81[ 0.1813 - 0.4226] = - 2.37 m/s²

Again, using equations of motion,

v² = u² + 2as

So in thus case u =0 and since it is moving down the plane, the displacement will be negative, thus

(Δx) = - 8.44m

So, v² = 2(-2.37)(-8.44) = 40

Thus, v = √40 = 6.323 m/s

found in the nucleus

has mass of one amu

massxgravity x heght=weightxheight= pe

4900=490xh

h=10