Work Power Energy

Here's a bit of work on this problem. Some of these results are a bit curious in relation to the proposed answers, so please comment if you see something wrong. Suppose the horizontal distance between the block and the pulley is $x$ and $\theta$ is the angle between the horizontal and the rope.

$tan(\theta) = \frac{h}{x}$

Derive an expression for the work done on the block in going between two values of $\theta$ .

Horizontal force:

$F_h = F cos(\theta)$

Differential work:

$dW = -F_h dx = -F cos(\theta) dx$

Implicitly differentiate the tangent expression to get an expression for $dx$ in terms of $d\theta$ .

$sec^2(\theta) \frac{d\theta}{dx} = \frac{-h}{x^2} = -\frac{tan^2 (\theta)}{h} \\ \frac{d\theta}{dx} = -\frac{sin^2 (\theta)}{h} \\ dx = -\frac{h \, d\theta}{sin^2 (\theta)} \\ dW = Fh \, \frac{cos(\theta)}{sin^2 (\theta)} d\theta$

Integrate to get the work done on the block:

$W = Fh \int_{\theta_1}^{\theta_2} \frac{cos(\theta)}{sin^2 (\theta)} d\theta \\ = Fh \int_{sin(\theta_1)}^{sin(\theta_2)} \frac{du}{u^2} = Fh \big(\frac{1}{sin(\theta_1)} - \frac{1}{sin(\theta_2)}\big)$

The work done between $A$ and $B$ is therefore:

$W_{AB} = Fh \big(\frac{1}{sin(30^\circ)} - \frac{1}{sin(53^\circ)}\big) \approx \boxed{0.74786 \, Fh}$

Interestingly, this is close to Option (a) but not exactly the same. Option (b) is clearly wrong.

I'm not exactly sure what (c) is asking for, so I'll skip to (d). Equate the work done on the block to the block kinetic energy.

$Fh \big(\frac{1}{sin(30^\circ)} - \frac{1}{sin(37^\circ)}\big) = \frac{1}{2} m \big(\frac{dx}{dt}\big)^2 \\ \implies \frac{dx}{dt} \approx 0.82263 \sqrt{\frac{Fh}{m}} \\ x = \frac{h}{tan(37^\circ)} \approx 1.327045 \, h$

The speed of point $P$ is equal to the rate of change of the length of rope hypotenuse on the left side of the pulley (call this $L$ ):

$L^2 = x^2 + h^2 \\ 2L \frac{dL}{dt} = 2x \frac{dx}{dt} \\ \frac{dL}{dt} = \frac{x}{L} \frac{dx}{dt} = \frac{x}{\sqrt{x^2+h^2}} \frac{dx}{dt}$

Evaluating gives:

$\frac{x}{\sqrt{x^2+h^2}} \frac{dx}{dt} = \frac{0.82263}{\sqrt{1 + \frac{1}{1.327045^2}}} \sqrt{\frac{Fh}{m}} \approx \boxed{ 0.656982 \sqrt{\frac{Fh}{m}}}$

Interestingly again, this is very close to the proposed answer for (d), but not quite the same.

ACD are the correct answers