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kirill115 [55]
3 years ago
15

A box moving on a horizontal surface with an initial velocity of 20 m/s slows to a stop over a time period of 5.0 seconds due so

lely to the effects of friction. What is the coefficient of kinetic friction between the box and the ground?
Physics
1 answer:
egoroff_w [7]3 years ago
8 0

Answer:

μ= 0.408 : coefficient of kinetic friction

Explanation:

Kinematic equation for the box:

a=\frac{v_{f} -v_{i} }{t} Formula( 1)

a= acceleration

v_i= initial speed =0

v_f= final speed= 20 m/s

t= time= 5 s

We replace data in the formula (1):

a=\frac{0-20}{5}

a= -\frac{20}{5}

a= - 4m/s²

Box kinetics: We apply Newton's laws in x-y:

∑Fx=ma : second law of Newton

-Ff= ma Equation (1)

Ff is the friction force

Ff=μ*N Equation (2)

μ is the coefficient of kinetic friction

N is the normal force

Normal force calculation

∑Fy=0  : Newton's first law

N-W=0   W is the weight of the box

N=W= m*g  : m is the mass of the box and g is the acceleration due to gravity

N=9.8*m

We replace N=9.8m in the equation (2)

Ff=μ*9.8*m

Coefficient of kinetic friction ( μ) calculation

We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):

-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation

9.8*μ=4

μ=4 ÷ 9.8

μ= 0.408

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2 years ago
How much is a city of a body when it covers 600m in 5 min?
Fofino [41]

Answer:

2m/s

Explanation:

5min × 60sec

=300

now,

600÷300

=2

3 0
3 years ago
Any magnetic properties occur _____________
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A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele
Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

3 0
3 years ago
(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

4 0
3 years ago
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