Physics problems...?

Question

I did like over a 100 exam review questions, heres a couple I couldn't figure out. Show equations and work?

A giraffe that stands 6m tall bites an entire branch off a tree. How long does it take that branch to hit the ground? What is the final velocity of the branch just before it hits the ground? What is its kinetic energy just before it hits the ground? What is the mass of the branch?

A rocket with a mass of 2 x 10^6 kg is launched with a velocity of 8000. how high does it rise?

A 4kg ball starts from rest and rools down a hill 3.5 m high and up an adjoining hill 2.5 m high. What will its speed be when it reaches the top of the second hill?

A 2 kg object initially at rest loses 400 J of potential energy while falling to the ground. Calculate the kinetic energy that the object gains while falling. What is the objects speed just before it striked the ground?

Answer 1

g = 9.8 m/s^2

1)

S = Vot + (at^2)/2

6 = 4.9t^2 => t ~= 1.10 s

V^2 = Vo^2 + 2aS

V^2 = 2*9.8*6 => V ~= 10.84 m/s

Just before hitting the ground, its velocity would be 10.84 m/s

K = m/2*V^2 = 58.8*m (m is its mass)

There's no way of knowing its mass...

2) V^2 = Vo^2 + 2aS (a = -g)

0 = 8000^2 - 19.6S

S = 3,265.30 km

3) The total amount of energy is constant.

Ei = mgh = 4*3.5*9.8 = 137.2 J

Ef = 4*2.5*9.8 = 98 J (only potential)

K = Ei - Ef = 39.2 J

m/2*V^2 = 39.2 = 2V^2 => V ~= 4.43 m/s

4) K = 400 J (the energy is constant, as there are no dissipative forces)

m/2*V^2 = 400 = V^2 => V = 20 m/s

__________________________

I replaced g. These results are not accurate enough for 0.2 m/s² to be a big deal, as the wind has not been taken into consideration. It's very inaccurate, in fact.

Answer 2

A giraffe that stands 6m tall bites an entire branch off a tree.

<< How long does it take that branch to hit the ground?>>

S =VT + (1/2)gT^2

where

S = 6 meters

V = initial velocity = 0

T = time

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

6 = 0 + (1/2)(9.8)T^2

T^2 = 2 * 6/9.8

T^2 = 1.224

T = 1.107 sec.

<<  What is the final velocity of the branch just before it hits the ground? >>

Potential energy = Kinetic energy

mgh = (1/2)mV^2

where

m = mass of branch

g = 9.8 m/sec^2

h = 6 m

V = velocity of branch as it hits the groung

Substituting values,

(m)(9.8)(6) = (1/2)(m)V^2

since "m" appears on both sides of the equation, it will simply cancel out, hence

V^2 = 2*9.8*6 = 117.6

V = 10.84 m/sec.

<< What is its kinetic energy just before it hits the ground? >>

KE before it hits the ground = (1/2)mV^2

KE = 58.8(m)

<<  What is the mass of the branch? >>

The mass of the branch cannot be determined from the data and related calculations of the above problem.

<< A rocket with a mass of 2 x 10^6 kg is launched with a velocity of 8000. how high does it rise? >>

Your working equation is

Vf^2 - Vo^2 = 2gs

where

Vf = final velocity = 0 (when rocket reaches its peak)

Vo = initial velocity = 8000 m/sec.

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

s = maximum height attained

Substituting values,

0 - 8000^2 = 2(-9.8)s

NOTE the negative sign attached to the acceleration due to gravity. This simply means that the rockets is slowing down as it goes up.

Solving for "s",

s = 3,265,306 meters = 3265.306 km.

<< A 4kg ball starts from rest and rools down a hill 3.5 m high and up an adjoining hill 2.5 m high. What will its speed be when it reaches the top of the second hill? >>

Potential energy loss = Kinetic energy gain

mg(3.5 - 2.5) = (1/2)mV^2

As in a previous problem, since "m" appears on both sides of the equation, it will simply cancel out.

9.8(1) = (1/2)V^2

V^2 = 2 * 9.8

V^2 = 19.6

V = 4.43 m/sec.

<< A 2 kg object initially at rest loses 400 J of potential energy while falling to the ground. Calculate the kinetic energy that the object gains while falling. What is the objects speed just before it striked the ground? >>

Conservation of energy --- potential enery loss = kinetic energy gain

Hence since potential energy loss = 400 J = Kinetic energy gain

and since KE = (1/2)mV^2

then

400 = (1/2)(2)(V^2)

V^2 = 400 * 2/2

V^2 = 400

V = 20m/sec.