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muminat
3 years ago
14

What must be the mass of a rock if a boy applies a 64N force and causes it to accelerate at 4.51m/s2

Physics
1 answer:
Butoxors [25]3 years ago
7 0

first of all the formula of force is F=ma,so we are searching for m,so we can divide a on both sides F/a=m, after this substitute the values given above 64N/4.51=14.2°Kg

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A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
types of plate boundaries where two plates separate or move apart?? does anyone know please help ??????
Shtirlitz [24]
Erosion i belive it is called
7 0
3 years ago
Read 2 more answers
In any energy transformation, there is always some energy that gets wasted as non-useful heat.
Nady [450]
It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
8 0
4 years ago
Read 2 more answers
This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi
Rufina [12.5K]

Answer:

 N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

            N - w₁ -w₂ =

            N = m₁ g + m₂ g

            N = g (m₁ + m₂)

let's calculate

            N = 9.8 (0.760 + 1.630)

            N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0
3 years ago
a force vector f has a magnitude of 12.0 n. it is oriented 60° to the left of the y ax what are its x and y components?
Arte-miy333 [17]

The  force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:

  • v(x) =6 N
  • v(y) = 10.39 N

To solve this exercise the formulas and procedures we will use are:

  • v(x) = v * cosine (angle)
  • v(y) = v * sine (angle).

Where:

  • v= magnitude of the vector
  • v(x) = component of the vector on the (x) axis
  • v(y) = component of the vector on the (y) axis
  • angle = angle

Information about the problem:

  • angle = 60º
  • v = 12.0 N
  • v(x)= ?
  • v(y)= ?

Applying the formula of the component of the vector in the (x) axis we have:

v(x) = v * cosine (angle).

v(x) = 12.0 N * cosine (60º)

v(x) =6 N

Applying the formula of the component of the vector in the (y) axis we have:

v(y) = v * sine (angle)

v(y) = 12.0 N * sine (60º)

v(y) = 10.39 N

<h3>What is a vector?</h3>

It can be said to be a straight line described by a point (a) and (b) that has direction and sense.

Learn more about vector at: brainly.com/question/2094736

#SPJ4

5 0
2 years ago
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