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4 years ago

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If the coefficient of static friction is 0.40, and the same ladder makes a 51° angle with respect to the horizontal, how far alo

ng the length of the ladder can a 57.0-kg person climb before the ladder begins to slip? The ladder is 7.5 m in length and has a mass of 21 kg.

1 answer:

inysia [295]4 years ago

5 0

To solve this problem, we apply the concepts related to the sum of forces and balance in a diagram that will be attached, in order to identify the behavior, direction and sense of the forces. The objective is to find an expression that is in terms of the mass, the angle, the coefficient of friction and the length that allows us to identify when the ladder begins to slip. For equilibrium of the ladder we have,

$\sum F_x = 0$

$\sum F_y = 0$

$\sum M_o = 0$

Now we have that

$f_1 = N_2$

$N_1 = mg$

And for equilibrium of the two forces we have finally

$mgdcos\theta = N_2lsin\theta$

Rearranging to find the distance,

$d = \frac{N_2}{mg}ltan\theta$

$d = \frac{f_1}{mg}ltan\theta$

So if we have that the frictional force is equivalent to

$f_1 = \mu N_1$

$f_1 = \mu mg$

$f_1 = (0.4)(57*9.8)$

$f_1 = 223.44N$

With this value we have that

$d = \frac{(0.4)(57)(9.8)}{57*9.8}(7.5) tan(60\°)$

$d = 5.19m$

Therefore can go around to 5.19m before the ladder begins to slip.

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

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Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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3 years ago

Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th

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Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components. While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

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4 years ago

Nitrogen gas at standard atmospheric pressure 101.3 kPa, has a volume of 0.080 m3. If there are 3.6 mol of the gas, what is the

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Explanation:

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3 years ago

Why are vectors important in physics?

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They are quantities with magnitude and direction like velocity, we'd be doomed without them to be honest , just knowing how fast something is going isn't enough, you want to know where it is going

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4 years ago

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

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$v^2=u^2+2as$

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

$0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s$

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

$v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s$

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

$v=u+at$

where symbols have the usual meaning

Applying the given values we get

$t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds$

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3 years ago