ll T-Mobile 3:27 PM * 39%. a session.masteringphysics.com C Chapter 2-3 Problem

Question

ll T-Mobile 3:27 PM * 39%. a session.masteringphysics.com C Chapter 2-3 Problem 2.58 A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.3 m. To jump this high, the bush baby accelerates over a distance of 0.16 m while extending the legs. The acceleration during the jump is approximately constant Part A What is the acceleration in m/s2 Express your answer using two significant figures Buent Request Answer Part B What is the acceleration in g's? Express your answer using two significant figures. Sna Request Answer Provide Feedback

Explanation / Answer

initial velocity = 0,
final velocity = v,
acceleration = a,
displacement = x
Therefore, v^2 = 0 + 2ax
Or v^2 = 2ax
Then kinetic energy at the time of jump is K1 = 1/2 mv^2
Kinetic energy at height h is K2 = 0 because in vertical jump, the speed at the heighest point = 0
Change in kinetic energy = K2 - K1 = 0 - 1/2 mv^2 = -1/2 mv^2
Change in potential energy = mgh
By conservation of mechanical energy,
change in kinetic energy + change in potential energy = 0
-1/2 mv^2 + mgh = 0
Dividing by m,
-1/2 v^2 + gh = 0
1/2 v^2 = gh
v^2 = 2gh
from the above two equation

2ax = 2gh

       a = gh/x

= 9.8 m/s2 * 2.3 m/0.16m

= 140.875 m/s2

Part B-

in g s

           h = 2.3 m - 0.16m =2.14m

a = g * 2.14m / 0.16m

= 13.375 g/s

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.