Question

A block of mass 0.107 kg is released from rest at point A, and it slides down a curved slope to point B. The vertical height of point A is h=2.06 m. What is the speed of the block at point B?

Answer 1

The speed of the block at point B is 6.36 m/s.

Step-by-step explanation:

To find the speed of the block at point B, we can use the principle of conservation of mechanical energy. At point A, the block only has potential energy which is given by mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the vertical height. At point B, the block has both potential energy and kinetic energy. Let's assume that the speed of the block at point B is vB.

Using the conservation of mechanical energy, we can write:

mgh = 0.5mvB2

Plugging in the given values, we have:

(0.107 kg)(9.8 m/s2)(2.06 m) = 0.5(0.107 kg)vB2

Solving for vB, we get:

vB = √(2gh)

Substituting the values, we find:

vB = √(2(9.8 m/s2)(2.06 m)) = 6.36 m/s