A block rests on a horizontal, frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an angle è above the horizontal, as shown in Figure 7-23. After the block is pulled through a distance of 1.50 m its speed is v = 2.90 m/s, and 50.0 J of work has been done on it.
what is the question?
To calculate the angle (θ) mentioned in the problem, we need to use trigonometry. The force mentioned (45.0 N) is the tension in the string, and we need to break it down into its horizontal and vertical components.
The horizontal component of the tension force (Fx) can be calculated using the formula:
Fx = T * cos(θ)
We know that Fx is responsible for the work done on the block. Given that 50.0 J of work has been done, we can equate it to the force and multiply by the distance traveled:
Work = Fx * distance
50.0 J = Fx * 1.50 m
Now, let's solve for Fx:
Fx = 50.0 J / 1.50 m
Now that we know the horizontal component of the force, we can solve for the angle θ.
θ = arccos(Fx / T)
Plugging in the known values:
θ = arccos(Fx / 45.0 N)
After finding the angle (θ), we can proceed with the other calculations mentioned in the problem.