A constant horizontal pull of 8.25 N drags a box along a horizontal floor through a distance of 19.1 m.
Q How much work does the pull do on the box?
Ans: i got the answer as W = 158 J
But i have problem with the second question?
Question: Suppose that the same pull is exerted at an angle above the horizontal. If this pull now does 60.0 of work on the box while pulling it through the same distance, what angle does the force make with the horizontal?
Ans. theta = _____ degree.
So now you know that W=60. So you set up the equation again using what you know.
x=theta
60=8.25*19.1*Cosx
Solve for x
60/157.575=cosx
cos^-1(.380)=x
x=67.67 degrees
Well, well, well, we've got ourselves a good old angle problem here! Don't worry, I'll make it as easy as pie... or should I say pi? Let's dive in!
Now, we know that the work done by a force is given by the formula W = F * d * cos(theta), where F is the force, d is the distance, and theta is the angle between the force and the displacement.
In the first question, we have the work W = 158 J and the distance d = 19.1 m. We also have the force F = 8.25 N pulling horizontally, so theta = 0 degrees (since the force is parallel to the displacement).
But in the second question, we are told that the same force F does 60.0 J of work. So now we have W = 60 J and d = 19.1 m. We need to find the angle theta.
We can rearrange the formula to solve for theta: theta = acos(W / (F * d))
Substituting the given values, we get:
theta = acos(60 / (8.25 * 19.1))
Now let me just crunch the numbers... *calculating sounds*... and voila! The angle theta is approximately _______ degrees! Fill in the blank, my friend, and you'll have your answer!
To find the angle that the force makes with the horizontal, we can use the work done on the box and the formula for work.
The work done on an object is given by the formula:
W = F * d * cos(theta)
Where:
W is the work done (given as 60.0 J),
F is the magnitude of the force (given as 8.25 N),
d is the distance the box is pulled (given as 19.1 m), and
theta is the angle between the direction of the force and the direction of the displacement.
Rearranging the formula, we get:
cos(theta) = W / (F * d)
Plugging in the values, we have:
cos(theta) = 60.0 J / (8.25 N * 19.1 m)
Now, let's calculate:
cos(theta) = 60.0 J / 157.275 Nm
cos(theta) = 0.381 (rounded to three decimal places)
To find theta, we need to take the inverse cosine (also known as the arccosine) of 0.381. Using a calculator or trigonometric table, we find:
theta ≈ 66.92 degrees
Therefore, the force makes an angle of approximately 66.92 degrees with the horizontal.
To determine the angle that the force makes with the horizontal, we can use the principles of work and energy. The work done by a force can be calculated using the formula:
Work = Force * Distance * cos(theta)
Where:
- Work is the work done by the force (in joules, J)
- Force is the magnitude of the force (in newtons, N)
- Distance is the distance over which the force acts (in meters, m)
- theta is the angle between the direction of the force and the direction of displacement (in degrees)
In this case, we're given that the work done by the force is 60.0 J, the distance is 19.1 m, and we want to find the angle theta.
Using the equation above, we can rearrange it to solve for theta:
theta = arccos(Work / (Force * Distance))
Substituting the given values into the equation:
theta = arccos(60.0 J / (8.25 N * 19.1 m))
Calculating this gives us:
theta = arccos(0.3807)
Using a calculator to find the arccosine of 0.3807, we get:
theta ≈ 67.7 degrees
Therefore, the angle that the force makes with the horizontal is approximately 67.7 degrees.