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dimulka [17.4K]

3 years ago

14

A 62 pound box is on an incline. Determine the minimum force P that would result in the box starting to slide up the incline. (i

nclude units with answer)

1 answer:

Anon25 [30]3 years ago

8 0

Answer:

F > W * sin(α)

Explanation:

The force needed for the box to start sliding up depends on the incline (α).

The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.

These forces can be decomposed on their normal and tangential (to the slide plane) components.

The weight will be split into

Wn = W * cos(α) (in normal direction)

Wt = W * sin(α) (in tangential direction)

The normal reaction will be alligned with the normal axis, and will be equal to -Wn

N = -W* cos(α) (in normal direction)

To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger

F > |-W * sin(α)| (in tangential direction)

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The carpet has a buildup of electrons in it that get released onto your body when you rub across the floor while the hardwood doesn't have electrons built up inside of it.

Explanation:

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2 years ago

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A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is

gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is $\theta = 25 ^o$

$sin \theta = \frac{h}{d}$

Where $h = h_b - h_a$

So $d sin \theta = h_b - h_a$

From the question we are told that

The mass of the block is $m = 20 \ kg$

The mass of the pendulum is $m_p = 2 \ kg$

The velocity of the pendulum at the bottom of swing is $v_p = 15 m/s$

The coefficient of restitution is $e =0.7$

The coefficient of kinetic friction is $\mu _k = 0.5$

The velocity of the block after the impact is mathematically represented as

$v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where $v_2 i$ is the velocity of the block before collision which is 0

$= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15$

Substituting value

$v_2 f = 2.310\ m/s$

According to conservation of energy principle

The energy at point a = energy at point b

So $PE_A + KE _A = PE_B + KE_B + E_F$

Where

$PE_A$ is the potential energy at A which is mathematically represented as

$PE_A = m_b gh_a$ = 0 at the bottom

$KE _A$ is the kinetic energy at A which is mathematically represented as

$K_A = \frac{1}{2} m_b * v_2f^2$

$PE_B$ is the potential energy at B which is mathematically represented as

$PE_B = m_b gh$

From the diagram $h = h_b -h_a$

$PE_B = m_b g(h_b - h_a)$

$KE _B$ is the kinetic energy at B which is 0 (at the top )

Where is $E_F$ is the workdone against velocity which from the diagram is

$\mu_k m_b g cos 25 *d$

So

$\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

Substituting values

$\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$

So

$d = 0.313 \ m$

6 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

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2 years ago

What is the density of a substance that has a mass of 453g and a volume of 224mL?

Mrrafil [7]

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3 years ago

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3 years ago

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