How much work is done by the horizontal force F = 150 N on the 18 Kg block when the force pushes the block 5.0 M up along the 32 degree friction-less incline?
F*(horizontal distance moved)
= 150*5.0*cos 32
= 636 J
You don't need the mass of the block to arrive at that answer. That is more than enough horizontal force to move the block up the frictionless ramp, so Kinetic Energy will increase.
I do get that answer also, but according to my textbook, the correct answer is
640J
Anyi, the 636J is rounded to 640J due to significant figures.
To calculate the work done by a force, we use the formula:
Work = Force * Distance * cosθ
In this case, the force (F) is given as 150 N, the distance (d) the block is pushed is 5.0 m, and the angle (θ) of the incline is 32 degrees.
To find the work done by the force, we need to calculate the component of the force that acts along the incline. This can be found by multiplying the force (F) by the cosine of the angle (θ):
Force along incline = F * cosθ
In this case, the force along the incline would be:
Force along incline = 150 N * cos(32°)
Next, we can substitute the values into the work formula:
Work = Force along incline * Distance
Work = (150 N * cos(32°)) * 5.0 m
Now, we can plug in these values to calculate the work:
Work = (150 N * 0.848) * 5.0 m
Work ≈ 637.2 J
Therefore, the work done by the horizontal force of 150 N on the 18 kg block is approximately 637.2 Joules.