Physics help needed

[Jacksaunt]

I'm taking physics one and I'm having some issues with a couple problems

157 joules is required to push a 4 kg box up a ramp 6 meters long.
How much force is required to push the box up the ramp at a constant velocity?

How much force is required to to lift the box straight up to a height of 4 meters?

Think of yourselves as the teachers

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[danisaac5]

Unless its a build up and release of energy, you already have the answer to your first question. If not (so its a constant force applied to the box), you would need a speed so you could work out the energy required per meter.

for question two, its two thirds of 157, so that would be roughly 104.66562 joules on a non-scientific calculator.

[Tom Gunn]

Just don't apply it

You could cause a Resonance Cascade

[Demian]

The problem didn't give any angles for the ramp? I can't think of a way to solve the first problem without any angles at the moment.


Here's the second problem of lifting the box up 4 meters.

Let's start by clearly defining everything we have.

Mass (m) = 4 kilograms (kg)
Height (h) = 4 meters (m)
Force (F) = ? Newtons (N)
Gravity of Earth (g) = 9.81 m/s²

I would use potential energy to solve this.

E_p = mgh

Let's define ground level as 0 potential. At 4 meters the box's potential energy is:

E_p = 4 kg * 9.81 m/s² * 4 m = 156.96 J

This means we are doing 156.96 J of work W to lift the box up 4 meters:

E_p = W

Work can be defined as:

W = Fh

Solve for F:

F = W / h = 156.96 J / 4 m = 39.24 N

I could be wrong. I haven't done physics in months. If you need it I could do some diagrams and expand the units.

[Ladezkik]

The 157 joules does not seem relevant to the second question, Dem.

Jacksaunt, are you sure the second question asked for amount of force rather than work? Are these the exact questions in the form your teacher/book gave them?

For the first problem

Work = Force * distance
W = F d

W = 157 J
d = 6 m
F = ? N

F = W / d
F = 157 J / 6 m ≈ 26.2 N

The angle of the ramp is irrelevant as you know the length of the path travelled and the work done by the object.

I'm dumb. You do need the angle of the ramp.

Try 2:

I'll just make the assumption the ramp is 6m long and 4m high and work based on that.

Step 1: Finding the angle of the ramp

Since

sin(Angle) = Opposite / Hypotenuse

Angle = arcsin ( Opposite / Hypotenuse )

Angle = arcsin ( 4 / 6 ) = arcsin ( 2 / 3 ) = 41.8 degrees

A = 41.8 deg

Step 2: Finding the component of the force of gravity along the ramp

For an object to be in equilibrium, in other words to be able to move at constant velocity, the forces acting on it need to add up to 0.

The force to be applied on the object will be upwards along the ramp. However, the force of gravity on the box is partially acting along the ramp as it is sloped. This 'component' of the force of gravity acts downwards along the ramp.

To have the box at equilibrium and allow it to move with constant velocity on the ramp, the forces acting on the box along the ramp need to add up to 0, meaning the force up the slope we need to apply on the box is equal in size to the component of gravity acting on it.

Since the angle of the ramp is 41.8 degrees, the component of the force of gravity acting along the ramp is given by

sin(41.8) = ( Component of gravity along the ramp / Force of gravity on the box )

sin(41.8) = ( F[g along ramp] / F[g] )

The force of gravity acting on an object F[g] is given by

F[g] = m * g

m = 4kg
g = 9.81 ms-2

F[g] = 39.2

Meaning the component of gravity along the ramp is

F[g along ramp] = F[g] * sin(41.8)

F[g along ramp] = 39.24 * 2/3 = 26.16 N

Since the force acting on the box up the ramp must be equal in size to the force acting on it down the ramp, the force needed to push it up with constant velocity is

F = 26.16 ≈ 26.2

which is uhhh .. equal to the value found with F = W / d ... lol

(at least it seems I was right in assuming that the ramp is 4m high)

[Fluff-is-back]

I'm dumb. You do need the angle of the ramp.

You don't. Your solution was correct assuming the force is parallel to the ramp.

A difference in angle will change the amount of work done. Since you are given the value, the angle is not required.

[SPooK]

Angle, gravity, mass and height are all irrelevant. Joules is a measurement of work done and is not the same as force. Force is only part of the equation used to get the amount of Joules. This is all really basic stuff guys, I don't know why you are ranting with long ass equations.

a) Work Done = Force x Distance

So 157 = Force(N) x 6 meters

Force in Newtons to push it up the ramp is 26.166 repeating.

b) If it is on the basis that you are using 157 joules to lift the box straight up as well then the formula is the same only instead of distance you use height.

So 157 = Force(N) x 4

Force in Newtons to pull it straight up is 39.25.

Took AP Physics for Junior year. For an understanding look at this image below.


 

[Smelly sock]

answer with "i don't know"

[Demian]

Angle, gravity, mass and height are all irrelevant. Joules is a measurement of work done and is not the same as force. Force is only part of the equation used to get the amount of Joules. This is all really basic stuff guys, I don't know why you are ranting with long ass equations.

a) Work Done = Force x Distance

So 157 = Force(N) x 6 meters

Force in Newtons to push it up the ramp is 26.166 repeating.

The part about constant velocity bothers me. For the velocity to be constant, F needs to be equal to G_x which is parallel to the ramp surface. You need the angle to figure out G_x. If the velocity is not constant then F is larger or smaller than G_x.

[Anti-Block]

itt: too many cooks spoil the batch