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3 years ago

5

A 65 kg student climbs 7 m up a rope at a constant speed. If the students power to output is 300 w, how long does it take the st

udent to climb the rope

1 answer:

tino4ka555 [31]3 years ago

3 0

g≈10 m/s²

F=G=mg=65*10=650 N

L=F*h=650*7=4550 J

P=L/t=>t=L/P=4550/300=15.16 s

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The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

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3 years ago

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp

kotegsom [21]

Answer:

m2 = 83.3 g

Explanation:

by conservation of momentum principle we have

$m_1v_{i1} + m_2v_{i2} = m_2v_{f2}$

as both sphere has same speed so $v_{i2} = v_{i1}$

$m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}$

from conservation of kinetic energy principle we have

$\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2$

$v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}$

$v_{f1} = v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}$

$\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}$

substituting this value in above equation to get m2 value

$m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}$

solving for m2 we get

$m2 = \frac{m_1}{3}$

m_1 = 250 g

$=\frac{250}{3}$

m2 = 83.3 g

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Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm

Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

$I_1\omega_1 = I_2\omega_2$

now here we have

$I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2$

$I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2$

$I_1 = 256$

Now when children come to the position of half radius

then we will have

$I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2$

$I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2$

$I_2 = 160$

now from above equation we have

$256 (20.0 rpm) = 160(\omega_2)$

$\omega_2 = 32 rpm$

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An 80- kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15m/s . Sort the fol

torisob [31]

Answer:

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

Explanation:

Given that,

Mass of the quarterback, m = 80 kg

Mass of the football, m' = 0.43 kg

Speed of the football, v' = 15 m/s

We need to sort the following quantities as known or unknown.

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

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