Can someone please help me with this. Thanks The plot shows the acceleration of

Question

Can someone please help me with this. Thanks

The plot shows the acceleration of a 5.22 kg particle while an applied force moves it from rest along 6 pi x - axis from x = 0 to x = 9.0 m. The limits of the scale of the vertical axis are set by as = 8.84 ms^-2. Why is the value the work done on the object by this force when it reaches x = 4.0 m? W = 161.5 J What is the object's speed it reaches x = 4.0 m? v = ms^-1 How much work has the force done on the particle when it reaches x = 7.0 m? What is the speed of the particle when it reaches x = 7.0 m? v = ms^- 1 How much work has the force done on the object when the object reaches x = 9.0 m? what is the object's speed when it reaches x = 9.0 m? v = ms^-1

Explanation / Answer

a) Area under the acceleration-position graph from x = 0 to x = 4m is

A = (1/2)(8.84m/s2)(4m+3m) = 30.94m2/s2

We can easily obtain the area under the Force-position graph from the area under acceleration-position graph by multiplying A by mass of the body.

So work done = area under the Force-position graph = mA = (5.22kg)(30.94m2/s2) = 161.5J

b) We invoke here the work-energy theorem which says that:

Work done on the system = change in kinetic energy of the system

or W = KEfinal - KEinitial

But the body starts from rest. So, KEinitial = 0

So KEfinal = (1/2)mv2, where v = speed of the particle at x = 4m

Now W = KEfinal = (1/2)mv2

or 161.5J = (1/2)(5.22kg)v2

or v = 7.866m/s

c) We repeat the procedure of (a)

Area under the acceleration-position graph from x = 0 to x = 7m is

A = (1/2)(8.84m/s2)(5m+3m) + (1/2)(1m + 2m)(-8.84m/s2)

or A = (1/2)(8.84m/s2)[(5m+3m) - (1m + 2m)] = 22.1m2/s2

Again, we can easily obtain the area under the Force-position graph from the area under acceleration-position graph by multiplying A by mass of the body.

So work done W = area under the Force-position graph = mA = (5.22kg)(22.1m2/s2) = 115.362J

d) again, at x = 0 speed of the particle was zero. So we have from work-energy theorem:

W =  KEfinal = (1/2)mv2, where v = speed of the particle at x = 7m

115.362J = (1/2)(5.22kg)v2

or, v = 6.648m/s

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....

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