Hi i really need
a 4 kg ball and a 3 kg block of wood were placed in an arrangement as shown above when the ball was released, it fell through a distance of 5 m, pulling the block of wood up the ramp. if the pulleys were well oiled and the surface of the ramp exerts a frictional force of 2 n, what is the
final speed of the sphere after falling through the 5m ?
(a) 5.00 ms?
(c) 6.89 ms
(b) 5.34 ms?
(d) 10.0 ms?

(B) 5.34 m/s

Explanation:

We can solve this with forces or with energy.

First, let's solve using forces.

The sum of the forces on the sphere in the y direction is:

∑F = ma

T − Mg = M(-a)

T = Mg − Ma

The sum of the forces on the block parallel to the incline is:

∑F = ma

T − F − mg sin θ = ma

Substituting and solving for a:

Mg − Ma − F − mg sin θ = ma

Mg − F − mg sin θ = (m + M) a

a = (Mg − F − mg sin θ) / (m + M)

a = (4×9.8 − 2 − 3×9.8×⅗) / (3 + 4)

a ≈ 2.79 m/s²

So the velocity after 5 meters of travel is:

v² = v₀² + 2a(x - x₀)

v² = (0)² + 2(2.79)(5)

v ≈ 5.29 m/s

Using energy:

Potential energy = Potential energy + Kinetic energy + Work

MgH = mgh + ½ Mv² + ½ mv² + Fd

(4)(9.8)(5) = (3)(9.8)(3) + ½ (4)v² + ½ (3)v² + (2)(5)

196 = 88.2 + 3.5 v² + 10

v ≈ 5.29 m/s

The closest answer is B, 5.34 m/s.

c, 1a hope this

vapour or gas assuming atmospheriv pressure