a block of mass 5.70 kg sits on top of an incline at an angle of 20.0 to the horizontal. the height of this object from the ground h= 1.10m

Respuesta :

Kinetic friction is the name given to the force that exists between moving surfaces.

Kinetic Friction: What is it?

  • The force that prevents an object from moving when it is sliding is known as friction. Because of kinetic friction, it is impossible for two or more things to move together at once.
  • The force acts in the direction that an object would prefer to slide, not the opposite.
  • Friction is produced when we apply the brakes to a moving vehicle in order to stop it. The Greek letter "mu" with the subscript "k" stands for the kinetic friction coefficient.
  • A body is subjected to k times the usual force due to kinetic friction. It is quantified in Newtons (N).
  • The following formula for calculating kinetic friction: (Kinetic friction coefficient) = (Kinetic friction force) (normal force).
  • The following sentences provide descriptions of the two main types of friction: Constant Resistance and Kinetic Friction.
  • The block's mass is 5.70 kg, and the incline's height is 1.10 meters. This is the gravitational potential.

(Complete Question: In Figure 853, a block with mass m=2.5 kg enters a spring with spring constant k=320 N/ m. By the time the block comes to a stop, the spring is crushed by 7.5 cm. Kinetic friction between floors and blocks has a coefficient of 0.25. What are the blockfloor system's increased thermal energy and the work the spring force performs when the block is in contact with the spring and being brought to rest? How quickly does the block move toward the spring?

  1. Solution: For x=0.075m and k=320N/m, Eq. 726 yields Ws =[tex]-\frac{1}{2} kx^{2} =-0.90J[/tex]. For later use, this is equivalent to what ΔU is opposite of.
  • (b) We find that FN = mg when we study forces, which implies that [tex]F_{k}= m_{k}F_{N}=m_{k}=mg[/tex].When d=x, Eq. 831 yields : Δ[tex]E_{th} =f_{k}d=m_{k}mgx=(0.25)(2.5)(9.8)(0.075)=0.46J[/tex].
  • (c) The initial kinetic energy is given as [tex]K_{i} =[/tex]Δ[tex]U+[/tex]Δ[tex]E_{th} =0.90+0.46=1.36J[/tex] which results in [tex]V_{i} =\sqrt{x2K_{i}/m } =1.0m/s[/tex]  )

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