constant force of 3.0 N to a projectile

[adrakdavra]

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Q 76
Starting from a resting position at the left end of the railgun, the armature applies a constant force of 3.0 N to a projectile with a mass of 0.06 kg. How long will it take for the projectile to move 1.0 m?
. 0.02 s
. 0.04 s
. 0.20 s
. 0.40 s

Why acc is not 10m/s2
I used d = 1/2 at2
I know you can use work = 1/2mvf2

 

[mehc012]

Q 76
Starting from a resting position at the left end of the railgun, the armature applies a constant force of 3.0 N to a projectile with a mass of 0.06 kg. How long will it take for the projectile to move 1.0 m?
. 0.02 s
. 0.04 s
. 0.20 s
. 0.40 s

Why acc is not 10m/s2
I used d = 1/2 at2
I know you can use work = 1/2mvf2

Acceleration is not 9.8m/s² because the railgun is horizontal (the acceleration is not due to gravity).

Instead, use
F = m·a
3.0N/0.06kg = a
a = 50m/s²
then follow it up with
x_f = x₀ + ½ at²
1m = ½ · 50m/s² · t²
t² = 1/25s²
t = 1/5s = 0.2s

 

[adrakdavra]

Thanks, but is it also that the force is applied and it is not a free fall

 

[mehc012]

Thanks, but is it also that the force is applied and it is not a free fall

Gravity is completely irrelevant to horizontal acceleration. The question you posted is concerned with left-right movement. Gravity is a downward acceleration and therefore does not affect left-right velocity.

 

[milski]

Q 76
Starting from a resting position at the left end of the railgun, the armature applies a constant force of 3.0 N to a projectile with a mass of 0.06 kg. How long will it take for the projectile to move 1.0 m?
. 0.02 s
. 0.04 s
. 0.20 s
. 0.40 s

Why acc is not 10m/s2
I used d = 1/2 at2
I know you can use work = 1/2mvf2

0.5mv² will give you the final speed but the speed during acceleration is not constant, so you cannot use that directly. Since the acceleration is constant, you can calculate the average speed, but that's just making things unnecessary complicated.

See mehc012 posted for the solution. The acceleration has nothing to do with free fall - why do you expect g to come in play here?

Thanks, but is it also that the force is applied and it is not a free fall

No idea what this means.

 

[adrakdavra]

I am sure it is an easy concept, but the left-right movement projectile is confusing

 

[mehc012]

I am sure it is an easy concept, but the left-right movement projectile is confusing

Vector quantities (Forces, velocities, accelerations, etc) can be separated into vertical (up/down) and horizontal (left/right) components. If you are trying to figure out the horizontal displacement, you look for horizontal components of the velocity, force, and acceleration vectors. Gravity always acts in the vertical (up/down) direction, and therefore does not contribute to horizontal movements.

In your example, the free body diagram of the object would indeed have a downward force applied by gravity - but because it is resting on the rail, it would have an upward normal force of equal magnitude, so the overall vertical acceleration, including the effects of gravity, would be zero.

However, in the horizontal direction, the only force is the 3.0N force from the left. This means that the overall horizontal acceleration is directly related to only that force by F_horizontal = m·a_horizontal

 

[adrakdavra]

Vector quantities (Forces, velocities, accelerations, etc) can be separated into vertical (up/down) and horizontal (left/right) components. If you are trying to figure out the horizontal displacement, you look for horizontal components of the velocity, force, and acceleration vectors. Gravity always acts in the vertical (up/down) direction, and therefore does not contribute to horizontal movements.

In your example, the free body diagram of the object would indeed have a downward force applied by gravity - but because it is resting on the rail, it would have an upward normal force of equal magnitude, so the overall vertical acceleration, including the effects of gravity, would be zero.

However, in the horizontal direction, the only force is the 3.0N force from the left. This means that the overall horizontal acceleration is directly related to only that force by F_horizontal = m·a_horizontal

Thanks sweetheart, try wearing a T shirt saying I can kill you with my brain around your Patients in the Hospital

 

[mehc012]

Thanks sweetheart, try wearing a T shirt saying I can kill you with my brain around your Patients in the Hospital

I'd be happy to, cuz it'd mean I succeeded at all this mess and actually HAD patients!

😛