Question

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
After being pulled 3.07 meters the crate has a velocity of 5.61 m/s. What is the angle (degrees) made by the rope with the horizontal?

Answer 1

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$