0.25 Phy1220: relationships among kinematics, newton's laws, vectors  (Page 3/5)

We will analyze several aspects of the state of the rocket at the end of each leg. We will also compare alternative ways of computing the state of the rocket.

Leg A

During this leg, the rocket is manually lifted from the ground to the platform. The rocket hasno potential or kinetic energy while on the ground, so it begins with zero mechanical energy.

An external force must be provided to lift the rocket from the ground to the platform in order to overcome the internal force of gravity. As an externalforce, this force is capable of changing its mechanical energy, which it does.

When the rocket has been lifted onto the platform, the mechanical energy of the rocket consists of its gravitational potential energy, which is equal tothe work done to lift it to the platform. The kinetic energy will be 0 at that point because the rocket isn't moving.

Weight of the rocket = m*g = 10kg*9.8m/s^2 = 98 newtons

Work = f * d = 98N * 15m = 1470 joules

State at the end of Leg A

Thus, the total mechanical energy possessed by the rocket at the end of Leg A is 1470 joules.

Leg B

During this leg, which begins when the rocket engine fires, the rocket flies straight up as a result of a constant upward force exerted by the rocket engine.

The net acceleration

For this leg, we need to determine the net acceleration that is applied to the rocket. The net accelerationconsists of the upward or positive acceleration due to the force of the rocket engine and the downward or negative acceleration of gravity.

Aup = 150N/10kg = 15 m/s^2

Ag = -9.8 m/s^2

Anet = 15 m/s^2 - 9.8 m/s^2 = 5.2 m/s^2

How far will the rocket go?

The initial velocity of the rocket is zero. Given that, you learned in an earlier module that the distance that the rocket will travel during the burn is

d = 0.5*Anet*t^2 = 0.5 * (5.2 m/s^2) * (10s)^2, or

d = 260 meters

260 meters straight up

In other words, when the rocket runs out of fuel at the end of the 10-second burn, the rocket has traveledstraight up by 260 meters. Given that it started 15 meters above the ground, it is at a height of 275 meters above the ground at that point in time.

Total mechanical energy

At that point in time, the total mechanical energy possessed by the rocket consists of its gravitational potential energy plus its kinetic energy.

The kinetic energy

To compute the kinetic energy, we need to know the velocity. You learned in an earlier module that we can computethe velocity as

v = Anet * t = (5.2 m/s^2) * 10s = 52 m/s

You also learned earlier that the kinetic energy is equal to

KE = 0.5 * m * v^2 = 0.5 * 10 kg * (52 m/s)^2 = 13520 joules

Gravitational potential energy

You learned in an earlier module that the gravitational potential energy of an object due to its height above the surface of the earth is equal to

PEg = m * g * h = 10 kg * (9.8m/s^2) * 275 m = 26950 joules

Note that this value is computed using the height above the ground and not the height above the platform, which is 260 meters.

The total mechanical energy

Thus, the total mechanical energy at this point in time is