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Ksenya-84 [330]

4 years ago

5

The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start

s from rest and the horizontal force F = 200 N, what is the magnitude of the velocity of the crate when it has moved 2 m?

1 answer:

Yanka [14]4 years ago

7 0

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction $\mu _s=0.24$

Coefficient of kinetic friction $\mu _k=0.22$

mass of crate $m=20\ kg$

Force applied $F=200\ N$

maximum static Friction $F_s=\mu _sN$

$N=mg$

$F_s=0.24\times 20\times 9.8$

$F_s=47.04\ N$

thus applied force is greater than Static friction therefore kinetic friction will come into play

$F_k=\mu _kN$

$F_k=0.22\times 20\times 9.8=43.12\ N$

net Force on crate $F-F_k=ma$

$a=\frac{200-43.12}{20}=7.84\ m/s^2$

Magnitude of velocity can be obtained by using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

$v^2-0=2\times 7.84\times 2$

$v=\sqrt{31.37}$

$v=5.60\ m/s$

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3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

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As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

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where the suffix i and f means initial and final respectively.

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$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

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So, this means:

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We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

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3 years ago

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3 years ago

A student decides to give his bicycle a tune up. He flips it upside down (so there’s no friction with the ground) and applies a

Alinara [238K]

Answer:

Tangential velocity = 10.9 m/S

Explanation:

As per the data given in the question,

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3 years ago

Ludmilka [50]

Answer:

3.92 N (straight down)

2.36 N (down, parallel to the plane)

Explanation:

The only force acting on the system is apparently that due to gravity. The mass of the system is the sum of the masses of the blocks, so is ...

0.20 kg + 0.20 kg = 0.40 kg

The force due to gravity is given by ...

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That force is acting straight down, so is at an angle with respect to the direction the blocks are constrained to move.

The force down the plane is (3.92 N)·sin(37°) ≈ 2.359 N . . . (down the plane)

_____

<em>Comment on the answer</em>

We have given two answers to the question, because in this frictionless system, the force acting normal to the plane of motion is irrelevant to the system dynamics. The question asked for the net force on the system, which is the force due to gravity, so we have given that magnitude also.

6 0

3 years ago