Question

13.A 10 kg monkey swings on a vine from a point which is 40 m above the jungle

floor to a point which is 15 m above the floor. If the monkey was moving at 2 m/s
initially, what will be his speed at the 15 m point?

Answer 1

v = 22.45 m/s

Step-by-step explanation:

You can use the Conservation of Energy formula here,

K₁ + U₁ = K₂ + U₂ or mgh₁ +
`(1)/(2)`mv²₁ = mgh₂ +
`(1)/(2)`mv²₂

We can then plug in our values

(10kg)(10m/s²)(40m) +
`(1)/(2)`(10kg)(2m/s²) = (10kg)(10m/s²)(15m) +
`(1)/(2)`(10kg)(v²)

4000 + 20 = 1500 +5v²

4020 = 1500 + 5v²

2520 = 5v²

504 = v²

`√(504)` = v

22.45 m/s = v

Answer 2

22.2m/s

Step-by-step explanation:

Given parameters:

Mass of monkey = 10kg

Height of descent = 25m

initial velocity = 2m/s

Unknown:

Speed at the point which is 15m

Solution:

So to solve this problem, we apply one of the kinematics equation:

V² = U² + 2gH

V is the unknown final speed

U is the initial speed

g is the acceleration due to gravity

H is the height of fall

V² = 2² + (2 x 9.8 x 25 ) = 494

V = 22.2m/s