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Mars2501 [29]
1 year ago
12

A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.

The student gives her backpack a quick kick at t = 0, imparting to it speed v and causing it to slide across the elevator floor. At time t, the backpack hits the opposite wall a distance L away from the student. Find the coefficient of kinetic friction μk between the backpack and the elevator floor. (Use any variable or symbol stated above along with the following as necessary: g.)
Physics
1 answer:
Anna007 [38]1 year ago
8 0

Answer:

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

N - mg = ma

so normal force on the block is given as

N = mg + ma

now the magnitude of kinetic friction on the block is given as

F_f = \mu N

F_f = \mu (mg + ma)

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

a = - \frac{F_f}{m}

a = -\mu_k(g + a)

now we know that bag hits the opposite wall at L distance away in time t

so we have

d = v t + \frac{1}{2}at^2

L = vt - \frac{1}{2}(\mu_k)(g + a) t^2

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

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2 years ago
A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
2 years ago
A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
tankabanditka [31]

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

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so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

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A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
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Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

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at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

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2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

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Sati [7]

Answer:

  h = v₀² / 2g ,      h = k/4g     x²

Explanation:

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Final point. Highest on the path

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As or no friction the energy is conserved  

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