How Does Projectile Motion Affect Internal Energy Changes?

[zumi78878]

Homework Statement

A 5-kg projectile is fired over level ground with a velocity of 200m/s at an angle
of 25◦ above the horizontal. Just before it hits the ground its speed is 150m/s.
Over the entire trip the change in the internal energy of the projectile and air is:
A. +19, 000 J
B. −19, 000 J
C. +44, 000 J
D. −44, 000 J
E. 0

I calculated the answer and i got 44000, but i don't know whether its (+) or (-).
the answer says its +44000 (C).

A 0.75-kg block slides on a rough horizontal table top. Just before it hits a horizontal ideal spring its speed is 3.5m/s. It compresses the spring 5.7 cm before
coming to rest. If the spring constant is 1200N/m, the internal energy of the block
and the table top must have:
A. not changed
B. decreased by 1.9J
C. decreased by 2.6J
D. increased by 1.9J
E. increased by 2.6J

I got the answer of 2.6J, but again, i don't know whether its (+) or (-).

the answer says its -2.6J (B)

33. A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy of the block is equal to the potential energy of the spring, the spring is compressed a distance of:
A. v√(m/2k)
B. (1/2)mv2
C. (1/4)mv2
D. (mv2)/(4k)
E. (1/4)√(mv/k)

my answer (not listed in choice?):
1/2mv2=1/2kx2
x = v√(m/k)

61. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer
radius, are each free to rotate about a fixed axis through its center. Assume the hoop is
connected to the rotation axis by light spokes. With the objects starting from rest, identical
forces are simultaneously applied to the rims, as shown. Rank the objects according to their
angular accelerations, least to greatest.

A. disk, hoop, sphere
B. hoop, disk, sphere
C. hoop, sphere, disk
D. hoop, disk, sphere
E. sphere, disk, hoop

answer is D, but why is that so? i know the hoop has the greatest rotational inertia, but how is I inversely proportional to angular acceleration?

is it because of Torque = I x ∞?

A disk starts from rest and rotates around a fixed axis, subject to a constant net torque. The work done by the torque during the second 5 s is ______ as the work done during the first 5 s.
A. the same
B. twice as much
C. half as much
D. four times as much
E. one-fourth as much

I don't understand this question at all, but answer is D.

 

[ehild]

Homework Statement

A 5-kg projectile is fired over level ground with a velocity of 200m/s at an angle
of 25◦ above the horizontal. Just before it hits the ground its speed is 150m/s.
Over the entire trip the change in the internal energy of the projectile and air is:
A. +19, 000 J
B. −19, 000 J
C. +44, 000 J
D. −44, 000 J
E. 0

I calculated the answer and i got 44000, but i don't know whether its (+) or (-).
the answer says its +44000 (C).

You need to know what "internal energy" means. Do you know?

How did you get that 44000 J?
The projectile loses mechanical energy, where does that energy go?
The book is correct.

A 0.75-kg block slides on a rough horizontal table top. Just before it hits a horizontal ideal spring its speed is 3.5m/s. It compresses the spring 5.7 cm before
coming to rest. If the spring constant is 1200N/m, the internal energy of the block
and the table top must have:
A. not changed
B. decreased by 1.9J
C. decreased by 2.6J
D. increased by 1.9J
E. increased by 2.6J

I got the answer of 2.6J, but again, i don't know whether its (+) or (-).

the answer says its -2.6J (B)

Again, it is the question, what is internal energy.

Think what happens when you rub your hands together when they are cold? Where does the mechanical energy go when an object is subjected to friction?

33. A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy of the block is equal to the potential energy of the spring, the spring is compressed a distance of:
A. v√(m/2k)
B. (1/2)mv2
C. (1/4)mv2
D. (mv2)/(4k)
E. (1/4)√(mv/k)

my answer (not listed in choice?):
1/2mv2=1/2kx2
x = v√(m/k)

You got x=v2 √(m/k), but you have to give x in terms of the original speed, v. How are v2 and v related? What is the speed if the kinetic energy is half the initial one?

61. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer
radius, are each free to rotate about a fixed axis through its center. Assume the hoop is
connected to the rotation axis by light spokes. With the objects starting from rest, identical
forces are simultaneously applied to the rims, as shown. Rank the objects according to their
angular accelerations, least to greatest.

A. disk, hoop, sphere
B. hoop, disk, sphere
C. hoop, sphere, disk
D. hoop, disk, sphere
E. sphere, disk, hoop

answer is D, but why is that so? i know the hoop has the greatest rotational inertia, but how is I inversely proportional to angular acceleration?

is it because of Torque = I x ∞?

If ∞ is the angular acceleration, yes. Denote it with alpha. τ(torgue)=I*α.

A disk starts from rest and rotates around a fixed axis, subject to a constant net torque. The work done by the torque during the second 5 s is ______ as the work done during the first 5 s.
A. the same
B. twice as much
C. half as much
D. four times as much
E. one-fourth as much

I don't understand this question at all, but answer is D.

How do you get the work in case of rotation? And how do you get the angular displacement in case of constant angular acceleration?

The book's answer is not correct again, I am afraid.

ehild

 

[zumi78878]

You need to know what "internal energy" means. Do you know?

How did you get that 44000 J?
The projectile loses mechanical energy, where does that energy go?
The book is correct.

Umm no i don't know what internal energy is, can you please explain?
I just did 1/2(5)(150²-200²) = -43750J
but why is the answer positive?

Again, it is the question, what is internal energy.

Think what happens when you rub your hands together when they are cold? Where does the mechanical energy go when an object is subjected to friction?

the energy turns into heat? but does it go into the surroundings?

and why is this answer negative and not positive?

You got x=v2 √(m/k), but you have to give x in terms of the original speed, v. How are v2 and v related? What is the speed if the kinetic energy is half the initial one?

ummmm i can't think of it :S

How do you get the work in case of rotation? And how do you get the angular displacement in case of constant angular acceleration?

The book's answer is not correct again, I am afraid.

ehild

work = F*r*d?

θ = 1/2(α)(Δt)²?

 

[ehild]

Umm no i don't know what internal energy is, can you please explain?
I just did 1/2(5)(150²-200²) = -43750J
but why is the answer positive?

the energy turns into heat? but does it go into the surroundings?

and why is this answer negative and not positive?

In 1), you calculated the difference between the final and initial kinetic energy. It is negative, so mechanical energy is lost. Yes, it turns into heat and it warms up the surroundings, but it warms up the projectile, too.
Internal energy of an object is the energy of the random motion of its molecules/atoms. The molecules have 1/2 mv^2 translational kinetic energy, but can have also rotational and vibrational energies. The average energy per degrees of freedom is k_AT, k_A is the Boltzmann constant and T is the absolute temperature.

When you rub your cold palms together, your work against friction increases the kinetic energy of the molecules of your hand, rises the internal energy, and you feel that your hands warmed up.

In the problem, the surroundings and the projectile or the block gained energy, equal to the lost mechanical energy. The question was the change of internal energy. It increased. So the sign is ...?
In problem (2) friction will increase the total internal energy again.

In problem 3, The new kinetic energy is equal to the potential energy, so how is it related to the initial kinetic energy? Energy is conserved.

work = F*r*d?

θ = 1/2(α)(Δt)²?

The work in case of rotational motion is Frθ (torque multiplied with angular displacement). The second equation is correct. Calculate the work during the first 5 seconds and then the work during the first 10 seconds.
What is the work then during the second 5-s interval?

ehild

 

[zumi78878]

In 1), you calculated the difference between the final and initial kinetic energy. It is negative, so mechanical energy is lost. Yes, it turns into heat and it warms up the surroundings, but it warms up the projectile, too.
Internal energy of an object is the energy of the random motion of its molecules/atoms. The molecules have 1/2 mv^2 translational kinetic energy, but can have also rotational and vibrational energies. The average energy per degrees of freedom is k_AT, k_A is the Boltzmann constant and T is the absolute temperature.

When you rub your cold palms together, your work against friction increases the kinetic energy of the molecules of your hand, rises the internal energy, and you feel that your hands warmed up.

In the problem, the surroundings and the projectile or the block gained energy, equal to the lost mechanical energy. The question was the change of internal energy. It increased. So the sign is ...?
In problem (2) friction will increase the total internal energy again.

so is the answer for #2 wrong? answer says that it decreases internal energy.

and mu prof just replied to my email, and he said that the ans for both ques is that internal energy decreases in both cases? now I am really confused... but your answer makes so much sense.

In problem 3, The new kinetic energy is equal to the potential energy, so how is it related to the initial kinetic energy? Energy is conserved.

the new kinetic energy is half of the initial kinetic energy?

OHHH i get it now.

i just use (1/2)(1/2mv²) = 1/2kx²

thanks so much

The work in case of rotational motion is Frθ (torque multiplied with angular displacement). The second equation is correct. Calculate the work during the first 5 seconds and then the work during the first 10 seconds.
What is the work then during the second 5-s interval?

work during first 5: (25/2)τ*α
work during last 5: (75/2)τ*α

its 3 times as much work? but there's no 3 in there...

 

[ehild]

Well, your prof might define the internal energy in different way. Ask him, what internal energy is.
As for the last question, the book looks wrong again.
See http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.htmlehild

 

[zumi78878]

Thanks so much! I feel like I know so much more now. Hope I do well on the exam tmr! :)

 

[ehild]

I hope your prof will think the same.

ehild