Physics, 24.03.2020 22:30 angie249

The student in item 1 moves the box up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25 degrees with respect to the incline and with the same 185 N force, what is the acceleration up the the ramp? Assume that the coefficient of kinetic friction is 0.27

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Answer from: Gennn43

The acceleration of the box up the the ramp is 0.044 m/s².

The given parameters;

inclination of the ramp, θ = 12⁰applied force, F = 185 Ninclination of the force, θ = 25⁰coefficient of friction, μ = 0.27mass of the box, m = 41.2 kg

The normal force on the box is calculated as follows;

$F_n + Fsin(\theta) - mg cos(\theta) = 0\\\\F_n = mgcos(\theta) - Fsin(\theta)\\\\F_n = (41.2 \times 9.8 cos(12)) \ - \ \ 185 \times sin(25)\\\\F_n = 316.75 \ N$

The net force on the box is calculated as follows;

$Fcos(\theta) - mgsin(\theta) - \mu F_n = ma\\\\185\times cos(25) \ - \ 41.2 \times 9.8\times sin(12) \ -0.27(316.75) = 41.2 a\\\\-1.8 \ = 41.2 a\\\\a = \frac{-1.8}{41.2} \\\\a = - 0.044 \ m/s^2$

Thus, the acceleration of the box up the the ramp is 0.044 m/s².

Learn more about motion of box on inclined plane: link

Answer from: kvekrcurf

Acceleration up the ramp = -2.745 m/ $s^{2}$

Explanation:

Given

box is pulled at angle Θ = $25^{o}$

Force applied F = 185N

coefficient of friction, μ $_{k}$ = 0.27

mass of the box m = 35 kg

We know that,

acceleration due to gravity g = 9.8 m/ $s^{2}$

horizontal component $F_{x}$ = F cosΘ = 185 * cos $25^{o}$ =167.67

vertical component $F_{y}$ = F sinΘ =185*sin $25^{o}$ = 78.18

Another vertical component is due to gravity $F_{g}$ , force in given by

$F_{g}$ = mg

= 35 x 9.8

= 343.35 N

Normal force $F_{n}$ = $F_{g}$ - $F_{y}$

= 343.35 - 78.18

= 265.17 N

Frictional force $F_{k}$ = $F_{n}$ * μ $_{k}$

= 265.17 * 0.27

= 71.596 N

To find acceleration, we know that,

force = mass x acceleration

acceleration = $\frac{force}{mass}$

Here force is the summation of frictional force and horizontal component of the applied force. These force act in opposite directions.

force = $F_{k}$ - $F_{x}$

= 71.596 - 167.67

= -96.074

acceleration = $\frac{-96.074}{35}$

= -2.745 m/ $s^{2}$

Answer from: Quest

force required to detach it is given by

$f = 4 * 10^{-8} n$

charges are taken to be fundamental charge

$q_1 = q_2 = 1.6 * 10^{-19}c$

now we have

$f = \frac{kq_1q_2}{r^2}$

$4 * 10^{-8} = \frac{9*10^9 * 1.6*10^{-19}* 1.6* 10^{-19}}{r^2}$

$r = 7.6 * 10^{-11} m$

so above is the minimum distance at which it will detach

Answer from: Quest

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The student in item 1 moves the box up a ramp inclined at 12 degrees with the horizontal. If the box...

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