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AlekseyPX
3 years ago
11

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

e displacement at 2 seconds.
Physics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

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Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is
marshall27 [118]

Answer:

x=0.0049\ m= 4.9\ mm

d=0.01153\ m=11.53\ mm

Explanation:

Given:

  • mass of the object, m=0.75\ kg
  • elastic constant of the connected spring, k=150\ N.m^{-1}
  • coefficient of static friction between the object and the surface, \mu_s=0.1

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

f=k.x

\mu_s.N=k.x

where:

N = m.g = the normal reaction force acting on the body under steady state.

0.1\times (9.8\times 0.75)=150\times x

x=0.0049\ m= 4.9\ mm

(b)

Now, according to the question:

  • Amplitude of oscillation, A= 0.0098\ m
  • coefficient of kinetic friction between the object and the surface, \mu_k=0.085

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

U=\frac{1}{2} k.A^2

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

U=F_k.d

\frac{1}{2} k.A^2=\mu_k.N.d

0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d

d=0.01153\ m=11.53\ mm

<em>is the total distance does it travel before stopping.</em>

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