Physics, 26.06.2019 07:30 CalCDanG

Adam and natalie push on a cart using 150 n of force. the cart does not move. what is static friction on the cart?

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Answer from: yolo3745

When force is applied on an object the object will not slide

So here object will remain in equilibrium after applying the force

So in this case the condition of equilibrium is maintained due to friction force that is applied on the object.

So here for equilibrium be can say

$F_{net} = 0$

also we know that

$F_{net} = F - F_f$

we are given that

F = 150 N ( applied force on the box)

$F_{net} = 150 - F_f = 0$

F_f = 150 N

So here this friction force on the cart is known as static friction

F_s = 150 N

Answer from: zawnghkawng64361

150N

Explanation:

Static friction is the friction that exists between two surfaces in contact. For a static body, the applied force (Fm) used to move the body is always equal to the frictional force(Ff) since frictional force is a force of opposition that tends to oppose the moving force acting on a static body.

According to the question, if Adam and Natalie push on a cart using 150N force, then the force of friction on the cart will be equal and opposite to the applied force and will also be equal to 150N

Answer from: Quest

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )

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.

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