Ask question

Ask question

All categories

3 years ago

11

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

0.53, respectively.
2
29◦
What is the frictional force acting on the 46 kg mass?

1 answer:

GREYUIT [131]3 years ago

7 0

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

#LearnwithBrainly

You might be interested in

A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient

Basile [38]

Answer:

v_o = 4.54 m/s

Explanation:

<u>Knowns </u>

From equation, the work done on an object by a constant force F is given by:

W = (F cos Ф)S (1)

Where S is the displacement and Ф is the angle between the force and the displacement.

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:

K.E=1/2m*v^2 (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:

W = K.E_f-K.E_o (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:

∑F_y=N-mg

N=mg

Thus, the kinetic friction force is:

f_k = μ_k*N

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:

W_f=(f_k*cos(180) s

=-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:

W= -K.E_o

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force

W_f= -K.E_o

From equation (2), the work done by the friction force in terms of the initial speed is:

W_f=-1/2m*v^2

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:

-μ_k*mg*s = -1/2m*v^2

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s

v_o = 4.54 m/s

6 0

3 years ago

Read 2 more answers

Fifty points please help?

Zina [86]

it would be C laminated soda lime glass

6 0

3 years ago

Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m

tatyana61 [14]

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

7 0

3 years ago

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0

3 years ago

A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an

Vlada [557]

Answer:

Explanation:

Given

volume $V=16 ft^3$

Suppose base is square with side L

height of crate is h

Volume $V=L^2\times h$

$16=L^2\times h$

Cost of top and bottom area $c_1=3L^2$

Cost of Side area $c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}$

Total Cost $C=c_1+c_2$

Total Cost $C=3L^2+\frac{128}{L}$

Differentiate C w.r.t Length

$\frac{dC}{dL}=6L-\frac{128}{L^2}$

$L^3=\frac{128}{6}$

$L=2.75 ft$

$h=\frac{16}{2.75^2}=11.46 ft$

Dimensions are $L\times L\times h=2.75\times 2.75\times 11.46$

6 0

3 years ago