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Vinvika [58]
3 years ago
14

Example: A wooden crate with mass 100kg is at rest on a stone floor. You know that the coefficients of kinetic and static fricti

on are: μs=0.5 and μk=0.4. A) What is the minimum horizontal force F need to just get the crate moving?
Physics
1 answer:
alexgriva [62]3 years ago
7 0

Answer

Any force greater 490N

Explanation

The force required just to make an object slide over a rough horizontal surface is any force greater that the static friction which given by;

F=\mu_s mg.............(1)

Given;

\mu_s=0.5\\m=100kg\\g=9.8m/s^2

Hence;

F = 0.5 x 100 x 9.8

F = 490N.

We will only need the coefficient of kinetic friction if we were asked to find the force required to keep the object moving uniformly. Usually, the force needed to keep an object moving uniformly over a rough surface is lesser that which is needed to start its motion.

In this problem, we were only asked to find the minimum force required to make the object move which we have done.

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Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

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3 years ago
Match the change of state of matter to the term:
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6 0
4 years ago
Read 2 more answers
Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a sta
yuradex [85]

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

7 0
3 years ago
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