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NeX [460]
2 years ago
7

in a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

ay. If the 3.5 kg book is pushed from rest through a distance of 0.96 m by the horizontal 22 N force from the broom and then has a speed of 1.36 m/s, what is the coefficient of kinetic friction between the book and floor?
Physics
1 answer:
madam [21]2 years ago
3 0

Answer:

μ=0.151

Explanation:

Given that

m= 3.5 Kg

d= 0.96 m

F= 22 N

v= 1.36 m/s

Lets take coefficient of kinetic friction = μ

Friction force Fr=μ m g

Lets take acceleration of block is a m/s²

F- Fr = m a

22 -  μ x 3.5 x 10 = 3.5 a         ( take g =10 m/s²)

a= 6.28 - 35μ  m/s²

The final speed of the block is v

v= 1.36 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.36² = 2 x a x 0.96

a= 0.963 m/s²

a= 6.28 - 35μ  m/s²

6.28 - 35μ = 0.963

μ=0.151

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4 0
2 years ago
A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50
stira [4]
Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward.   Hold that thought, while I slog through
the complete solution of the problem.


            (100 N forward) plus (50 N backward)

        =  (100 N forward) minus (50 N forward)

        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

4 0
2 years ago
What is the y component of a vector defined as 12.2m at 81.5°?
sergejj [24]

Answer:

Explanation:

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V_y=Vsin\theta which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

V_y=12.2sin(81.5) and get

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3 0
2 years ago
calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s
Black_prince [1.1K]

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

1.91J = I/2*\omega^2

where I = 2/3*m*R^2=0.23kg.m^2

Solving for ω:

\omega = \sqrt{W*2/I} =4.07rad/s

7 0
2 years ago
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