Physics, 24.03.2020 18:02 SoWhat2237

A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at an angle of 25 above the horizontal. The box has a mass of 35.0kg and a the coefficient of kinetic frictino between the box and the floor is 0.27. Find the acceleration of the box

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Answer from: LilCookies1

The acceleration of the box is $2.75 \;\rm m/s^{2}$ .

Given data:

The magnitude of force applied by the student is, F = 185 N.

The angle with horizontal is, $\theta = 25 ^{\circ}$ .

The mass of box is, m = 35.0 kg.

The coefficient of kinetic friction between box and floor is, $\mu = 0.27$ .

The given problem is based on the equilibrium of force. And to solve the problem, we need to obtain the equation of motion for horizontal and vertical component of force. So,

Along the horizontal direction,

$Tcos\theta -\mu N =ma$ .....................................................(1)

Here, $Tcos \theta$ horizontal component of tension force on the rope, N is the normal reaction force and a is the magnitude of acceleration of box.

Along the vertical direction, the forces are,

$N+Tsin\theta =mg\\\\N = mg -T sin\theta ...................................................................(2)$

Substitute the value of equation (2) in (1) as,

$Tcos\theta -\mu ( mg -T sin\theta )=ma\\\\a = \dfrac{Tcos\theta -\mu mg + T sin\theta }{m}$

The tension force on the rope is due to the applied force , T = F = 185 N.

Solving as,

$a = \dfrac{185 \times cos25 -(0.27) \times 35 \times 9.8 +( (185) sin25) }{35}\\\\a = 2.75 \;\rm m/s^{2}$

Thus, we can conclude that the acceleration of the box is $2.75 \;\rm m/s^{2}$ .

Learn more about the equilibrium of force here:

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Answer from: cannaincorporated

2.75 m/s^2

Explanation:

We can solve the problem by writing the equations of motion along the horizontal and vertical direction.

Along the horizontal direction we have:

$T cos \theta - \mu N = ma$ (1)

where

$T cos \theta$ is the horizontal component of the tension, where

T = 185 N is the tension in the rope

$\theta=25^{\circ}$ is the angle between the rope and the horizontal

$\mu N$ is the force of friction, where

$\mu=0.27$ is the coefficient of friction

N is the normal reaction of the floor on the box

m = 35.0 kg is the mass of the box

a is the acceleration

Along the vertical direction we have:

$N+T sin \theta-mg=0$ (2)

where

N is the normal force (upward direction)

$T sin \theta$ is the vertical component of the tension in the rope (upward direction)

is the weight of the box (downward direction), where

m = 35.0 kg is the mass of the box

g=9.8 m/s^2 is the acceleration due to gravity

From eq.(2) we get:

$N=mg-T sin \theta$

And substituting into (1), we can find the acceleration:

$T cos \theta - \mu (mg-T sin \theta) = ma\\Tcos \theta -\mu mg + \mu T sin \theta = ma\\a=\frac{T cos \theta- \mu mg + \mu T sin \theta}{m}=\\=\frac{(185)(cos 25^{\circ})-(0.27)(35.0)(9.8)+(0.27)(185)(sin 25^{\circ})}{35.0}=2.75 m/s^2$

Answer from: Quest

magnitude and direction

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

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