Question

Figure shows the vertical section of frictionless surface. A block of mass 2kg released from the position A; its KE as it reached point C is:

Answer 1

Answer:

hello

Explanation:

hope it's help u thnku .

Answer 2

Answer :

Answer is 140 Joules.

We know that,

• Block was the height of 14 m at A.
• And at the height of 7m at C.
• Mass of body = 2 kg.

Vertical height (h) = 14m - 7m

= 7m

According the law of conservation of energy,

• The potential energy of a body of height is converted into kinetic energy when it falls down.
• So, the magnitude of kinetic energy of block just before hitting the ground is equal to the potential energy at height which is MGH.

Therefore,

• Kinetic energy at C = MGH
• 2kg x 10m/s x 7m
• 140 Joules.

Hence, the required answer is 140 Joules.