Physics, 27.09.2019 19:30 HolliAnne

Abox is pushed down at an angle of 32 degrees on a rough surface. the box moves to the right. what equation should be used to find the net force in the y-direction?

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Answer from: carleygalloway103

$F_{net}=\mu mg cos 32^{\circ} sin 32^{\circ} +mg cos^2 32^{\circ} - mg - F sin 32^{\circ}$

Explanation:

We have four forces acting on the box. Taking the upward direction as positive direction for the vertical axis:

- The push, which pushes down along the ramp inclined at 32 degrees. Its vertical component will be $-F sin 32^{\circ}$ , where F is the magnitude of the push

- The weight of the box, of magnitude -mg , with m being the mass of the box and g being the acceleration of gravity (9.8 m/s^2). This force is already directed vertically downward, so we don't need to resolve it into the vertical direction

- The frictional force, of magnitude $\mu mg cos 32^{\circ}$ , where $\mu$ is the coefficient of friction, directed upward along the inclined plane. Its vertical projection will be $\mu mg cos 32^{\circ} sin 32^{\circ}$

- The normal reaction, whose magnitude is equal to the component of the weight perpendicular to the inclined plane, but with opposite direction (upward): $N=mg cos \theta$ . If we resolve it along the vertical direction, we get $N_y = mg cos 32^{\circ} cos 32^{\circ} = mg cos^2 32^{\circ}$

So, the net force along the y-direction will be

$F_{net}=\mu mg cos 32^{\circ} sin 32^{\circ} + mg cos^2 32^{\circ} - mg - F sin 32^{\circ}$

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

i believe the answer is electric potential.

explanation:

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Abox is pushed down at an angle of 32 degrees on a rough surface. the box moves to the right. what e...