a 17.0kg crate is to be pulled a distance of 20.0m requiring 1210J of work being done. If the job is done by attaching a rope and pulling with a force of 75.0 N, at what angle is the rope held?
W=Fd(cos)
1210J=(170)(20m)(cos)
To find the angle at which the rope is held, we can rearrange the formula for work (W = Fd cosθ) to solve for cosθ.
Given:
Weight of the crate (W) = 17.0 kg
Distance (d) = 20.0 m
Work required (1210 J)
Force (F) = 75.0 N
Let's substitute the values into the equation:
1210 J = (17.0 kg)(20.0 m)(cosθ)
To solve for cosθ, divide both sides of the equation by [(17.0 kg)(20.0 m)]:
1210 J / [(17.0 kg)(20.0 m)] = cosθ
The units of kg and m cancel out, leaving us with:
1210 J / (17.0 kg × 20.0 m) = cosθ
Calculating this expression, we get:
cosθ = 0.355
Now, to find the angle θ, we take the inverse cosine (arccos) of cosθ:
θ = arccos(0.355)
Using a calculator, the inverse cosine of 0.355 is approximately 69.38 degrees. Therefore, the angle at which the rope is held is approximately 69.38 degrees.
To find the angle at which the rope is held, we can use the equation:
W = Fd cos(theta)
Given:
W = 1210 J
F = 75.0 N
d = 20.0 m
Substituting the given values into the equation, we have:
1210 J = (75.0 N)(20.0 m) cos(theta)
Now, let's solve for cos(theta):
cos(theta) = (1210 J) / ((75.0 N)(20.0 m))
cos(theta) = 1210 J / 1500 Nm
cos(theta) ≈ 0.80667
To find theta, we can take the inverse cosine (cos^(-1)) of 0.80667:
theta ≈ cos^(-1)(0.80667)
theta ≈ 39.93 degrees (approx.)
Therefore, the angle at which the rope is held is approximately 39.93 degrees.