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Tatiana [17]
3 years ago
12

A massive launcher sends a projectile vertically upwards from the surface of a planet of mass M and radius R. You may assume tha

t this planet has no atmosphere. Part A If the projectile is launched at the escape speed, how do the magnitudes of its initial KE and gravitational potential energy (GPE) compare
Physics
1 answer:
uranmaximum [27]3 years ago
5 0

Answer:

It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.

Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.

This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.

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You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out
kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0
3 years ago
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