At the end of the semester, you are loading your gear into a van to go home. In loading the van, you can either lift a 45.0-kg crate straight up(method 1) or push it up a 3.00-m-long ramp that is inclined at an angle above the horizontal (method 2).In both methods, the crate is moved at a constant speed and the vertical distance from the ground to the floor of the van has the same value h. In method 1, no friction is present. In method 2, however, a frictional force does -485J of work on the crate. In applying your nonconservative pushing force to the crate, you do twice as much work in method 2 as in method 1. Find the angle.

I've posted this question, and someone gives me this answer. But I don't quite get it, can anyone please explain it to me?Thanks!

First, compute the work done in method 1. That would be M g H = 45*9.8*3.0 sin A = 1323 sin A Joules. A is the angle of the place to the horizontal.

With the ramp, you do twice as much work, or 2646 J sin A. If 485 J of this was overcoming friction, the potential energy change was
2646 sin A - 485
This equals the potential energy change with method 1, or 1323 sin A. Therefore
2646 sin A -485 = 1323 sin A
485 = 1323 sin A
sin A = 0.367
A = ?

The way I see if we could lift the crate straight up in method 1, no ramp.

We know what h is. It is 3 meters * sin A

Then the work done = m g h = 3 m g sin A

In method 2 we do twice as much work as in method 1. However the work we do against gravity in method 2 remains mgh so the work done against friction is also mgh

the work done against friction is the same as mgh so
3 m g sin A = 485 J
3(45)(9.8) sin A = 485
sin A = .367
A = 21.5 degrees

Looks like you're trying to understand the solution to a physics problem. Let me break it down for you in a funny way!

So, in method 1, you're lifting the crate straight up with no friction involved. It's like lifting weights at the gym, but without all the grunting and sweating.

The work done in method 1 can be calculated using the formula M g H, where M is the mass of the crate, g is the acceleration due to gravity, and H is the height. In this case, M = 45 kg, g = 9.8 m/s^2, and H = 3.0 m sin A (because the ramp is inclined at an angle A).

Now, let's talk about method 2, which involves pushing the crate up the ramp. It turns out that you're doing twice as much work compared to method 1. That's like getting a bonus for your hard work, but without the extra paycheck.

So, the total work done in method 2 is 2646 J sin A. But hold on, there's a sneaky frictional force of -485 J working against you. It's like having a mischievous little gremlin trying to ruin your day.

The potential energy change, which is equal to the work done, can be calculated using the formula 2646 sin A - 485. We want this to be equal to the potential energy change in method 1, which is 1323 sin A.

So, setting up an equation: 2646 sin A - 485 = 1323 sin A. We can solve this equation to find the value of sin A.

And after some calculations, we find sin A = 0.367. And that means A = ? (the angle we're looking for).

So, in a nutshell, the angle is the solution to the equation 485 = 1323 sin A.

I hope that explanation puts a smile on your face, even if the math doesn't always do the trick!

To solve the question, let's start by understanding the different components involved:

Method 1:
In method 1, the crate is lifted straight up with no friction. The work done in lifting the crate can be calculated using the formula: work = force × distance.

In this case, the force required to lift the crate is its weight, which is given by the formula: force = mass × gravity. The work done in method 1 is equal to the potential energy change of the crate.

Method 2:
In method 2, the crate is pushed up a ramp that is inclined at an angle above the horizontal. There is a frictional force present, which does work on the crate. The total work done in method 2 is equal to the work done by the pushing force plus the work done by the frictional force.

Given that the work done in method 2 is twice the work done in method 1, we can set up the equation:

Total work done in method 2 = 2 × work done in method 1

Let's substitute the values and solve for the angle (A):

Total work done in method 2 = 2646 J sin A
Work done in method 1 = 1323 J sin A (Using the formula: M g H = 45 * 9.8 * 3.0 sin A)

2646 J sin A = 2 × (1323 J sin A)

2646 sin A = 2646 sin A - 485

485 = 1323 sin A

sin A = 485 / 1323

Using a calculator, sin A ≈ 0.367

To find the angle (A), take the inverse sine of 0.367:

A = sin^(-1)(0.367)

Using a calculator, A ≈ 21.5 degrees.

Therefore, the angle (A) is approximately 21.5 degrees.

To solve this question, we need to compute the work done in both method 1 and method 2, and then set them equal to each other since the potential energy change should be the same in both methods.

Method 1: In this method, we are lifting the crate straight up without any friction. The work done is given by the formula W = M * g * H, where M is the mass of the crate, g is the acceleration due to gravity, and H is the vertical distance the crate is lifted.

In this case, the mass of the crate is 45.0 kg, the acceleration due to gravity is 9.8 m/s^2, and the vertical distance is 3.00 m. Therefore, the work done in method 1 is:
W1 = 45.0 * 9.8 * 3.00 * sin(A) Joules, where A is the angle of the ramp to the horizontal.

Method 2: In this method, the crate is pushed up a ramp with an inclined angle A above the horizontal. We are given that the work done in method 2 is twice as much as in method 1, so W2 = 2 * W1.

We are also given that a frictional force does -485 J of work on the crate in method 2. This means that the total work done in method 2 is the work done to overcome friction plus the work done to move the crate vertically.

Therefore, W2 = 2646 * sin(A) J, where 2646 is the total work done, and sin(A) represents the vertical component of the ramp.

We can set up the equation: 2646 * sin(A) = 485 + 1323 * sin(A), since the potential energy change in both methods should be the same.

Simplifying the equation, we get: 2646 * sin(A) - 1323 * sin(A) = 485

Combining like terms, we have: 1323 * sin(A) = 485

To isolate sin(A), we divide both sides by 1323: sin(A) = 485 / 1323

Using a calculator or trigonometric table, we find that sin(A) is approximately equal to 0.367.

To find the angle A, we can take the inverse sine (arcsin) of 0.367 using a calculator or trigonometric function, which gives us approximately 21.78 degrees.

Therefore, the angle A is approximately 21.78 degrees.