Physics, 05.08.2019 21:20, alexanderavrett

Astudent decides to move a box of books into her dormitory room by pulling on a rope attached to the box. she pulls with a force of 80.0 n at an angle of 25.0° above the horizontal. the box has a mass of 25.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (a) find the acceleration of the box. (b) the student now starts moving the box up a 10.0° incline, keeping her 80.0 n force directed at 25.0° above the line of the incline. if the coefficient of friction is unchanged, what is the new acceleration of the box?

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answered: kayranicole1

The acceleration of the box when the student was it along the horizontal is 0.37 m/s².

The acceleration of the box when it was pulled up 10⁰ incline is -1.29 m/s².

The given parameters;

mass of the box, m = 25 kgforce applied on the box, F = 80 Nangle of inclination of the rope, θ = 25 ⁰

The normal force on the box is calculated as;

$F_n = W - Fsin(\theta)\\\\F_n = (25\times 9.8) - (80 sin(25))\\\\F_n = 211.19 \ N$

The frictional force on the box is calculated as;

$F_k = \mu_k F_n\\\\F_k = 0.3\times 211.19 \\\\F_k = 63.36 \ N$

The acceleration of the box is calculated as;

$Fcos(\theta) - F_k = ma\\\\a = \frac{Fcos(\theta) - F_k}{m} \\\\a = \frac{80\times cos(25) - 63.36}{25} \\\\a = 0.37 \ m/s^2$

When the student starts pulling the box at 10.0° incline;

The normal force on the box is calculated as;

$F_n = Wcos(\theta) - Fsin(\theta)\\\\F_n = (25\times 9.8 \times cos(10)) - (80 sin(25))\\\\F_n = 207.27 \ N$

The frictional force on the box is;

$F_k = \mu_k F_n\\\\F_k = 0.3\times 207.27 \\\\F_k = 62.18 \ N$

The acceleration of the box is calculated as;

$Fcos(25) - F_k - Wsin(10) = ma\\\\(80cos(25) ) - 62.18 - (25\times 9.8\times sin(10)) = 25a\\\\-32.23 =25a\\\\a = \frac{-32.23}{25} \\\\a = -1.29 \ m/s^2$

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answered: dbegay36

Part a)

a = 0.36 m/s^2

Part b)

a = -1.29 m/s^2

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

F_x = Fcos25

F_x = 80 cos25 = 72.5 N

F_y = Fsin25

F_y = 80 sin25 = 33.8 N

now in vertical direction we can use force balance for the box to find the normal force on it

F_n + F_y = mg

F_n = (25)(9.81) - 33.8

F_n = 211.45 N

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

F_k = (0.300)(211.45) = 63.44 N

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

F_x = 72.5 N

F_y = 33.8 N

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

F_n = (25)(9.81)cos10 - 33.8 = 207.7 N

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

f_k = (0.300)(207.7) = 62.3 N

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

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Astudent decides to move a box of books into her dormitory room by pulling on a rope attached to the...