- #1
greenrichy
- 11
- 0
- Homework Statement
- A person drags a [itex]75 kg[/itex] block using a rope over a rough surface that is inclined [itex]\theta=41^{\circ}[/itex] above the horizontal (the block's initial velocity is zero). The block is dragged a distance of [itex]30.5 m[/itex] along the incline with a constant acceleration of [itex]0.25\frac{m}{s^2}[/itex]. The coefficient of kinetic friction between the block and the ground is [itex]0.31[/itex].
(a) Find the total work done by the person who is pulling the block (you may assume that the rope is parallel with the surface during the drag)
(b) The instant before the block reaches the top of the hill, it snaps. The block slides back down the hill. Determine the block's velocity the instant it reaches the bottom of the hill.
- Relevant Equations
- $$\sum F_x = T - w_x - f_k = ma_x $$
$$\sum F_x = T - w_x - f_k = ma_x $$
$$ T = mg\sin(\theta) + mg\cos(\theta)\mu_k + ma_x$$
$$ T = (9.8 \frac{m}{s^2}) \cdot (\sin(41^{\circ}) + \cos(41^{\circ})) + (75kg)\cdot(0.25\frac{m}{s^2}) $$
$$T = 672.91 N $$
Having found the tension force, I can find the work done by the person who's pulling the block:
$$ W = 672.91N \cdot 30.5m $$
$$ W = 20510.2968J $$
Quick question: can I solve the above part with the Work-Energy theorem?
Now, when it comes to part (b), I've solved it using the Work-Energy theorem:
$$ W = \Delta K $$
$$ (-f_k - w_x)\cdot \cos(0^{\circ}) \cdot (-d) = \frac{1}{2} m\nu^2$$
$$ \nu=\sqrt{2g(\cos(\theta)\mu_k + \sin(\theta))\cdot d} $$
After substituting values, I got the following answer for the block's final speed:
$$ \nu = 23.0587179 \frac{m}{s^2} $$
But how do I express the object's final velocity? Would the following suffice?
The object travels with a speed of [itex]23.1 \frac{m}{s^2}[/itex] at an angle of [itex]41^{\circ}[/itex].
$$ T = mg\sin(\theta) + mg\cos(\theta)\mu_k + ma_x$$
$$ T = (9.8 \frac{m}{s^2}) \cdot (\sin(41^{\circ}) + \cos(41^{\circ})) + (75kg)\cdot(0.25\frac{m}{s^2}) $$
$$T = 672.91 N $$
Having found the tension force, I can find the work done by the person who's pulling the block:
$$ W = 672.91N \cdot 30.5m $$
$$ W = 20510.2968J $$
Quick question: can I solve the above part with the Work-Energy theorem?
Now, when it comes to part (b), I've solved it using the Work-Energy theorem:
$$ W = \Delta K $$
$$ (-f_k - w_x)\cdot \cos(0^{\circ}) \cdot (-d) = \frac{1}{2} m\nu^2$$
$$ \nu=\sqrt{2g(\cos(\theta)\mu_k + \sin(\theta))\cdot d} $$
After substituting values, I got the following answer for the block's final speed:
$$ \nu = 23.0587179 \frac{m}{s^2} $$
But how do I express the object's final velocity? Would the following suffice?
The object travels with a speed of [itex]23.1 \frac{m}{s^2}[/itex] at an angle of [itex]41^{\circ}[/itex].