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jarptica [38.1K]
3 years ago
5

A 1650 kg car accelerates at a rate of 4.0 m/s^2. How much force is the car's engine producing?

Physics
2 answers:
Westkost [7]3 years ago
7 0
F = M A

Force = (mass) x (acceleration) 

                   = (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>
Eduardwww [97]3 years ago
3 0
According to Newtons laws of motion F = MA. This means Force = Mass x Acceleration.

So...

4 x 1650 = 66,000
The units for Force is Newtons

So the answer is 66,000 Newtons
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3 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
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Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
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