Question

The coordinate of a 2.00kg object in linear motion is described by the function:

x(t)= (2.00m/s^3)t^3 -(7.00m/s^2)t^2 + (7.00m/s)t

for the time interval t= 0s to t= 2.00s. The center of mass of this object can be treated as a point particle.

a) find change in momentum of the mass for the time interval given.

b) sketch graphs of the momentum of the object and the force on the object as functions of time

c) How much power is delivered to the particle at any time t?

d) How much work is done on the object fo the time interval given?

Does it matter that this is treated as a point particle at it's center mass?

[B]a) to get change in momentum [/B]

v(t)= x'(t)= (6.00m/s^3) t^2 - (14.00m/s^2) t + 7.00m/s

ti= 0s

v(0)= 7.00m/s

tf= 2.00s

v(2.00)= (6.00m/s^3)(2)^2 - (14.00m/s^2) (2) + 7.00m/s

v(2.00)= 2.00m/s

I= Pf- Pi = m(vf-vi)= 2.00kg (3.00m/s - 7.00m/s)

I= -8 kg*m/s ==> is this alright??

b) graph

how would I sketch graphs of the momentum of the object and the force on a object as a function of time?

Once again I'm not sure but I think this is how I'd do this...

to get momentum as a function of time...I think I'd multiply the mass into the velocity equation:

p= mv

v(t)= (6.00m/s^3) t^2 - 14.00m/s (t) + 7.00m/s

m= 2.00kg

[B]p(t)= (12.00kg*m/s ^3) t^2 - 28.00m/s^2 (t) + 7.00m/s[/B]


For the force as a function of time I think I'd have to multiply the mass * acceleration

a(t)= v'(t)= (12.00m/s^3) t- (14.00m/s^2)

F= m*a

m= 2.00kg

F(t)= (24.00kg*m/s^3) t - (28 kg*m/s^2)

[B]c) How much power is being delivered to the particle at any time t?[/B]

Not sure..

P= F*V

would I go and multiply the force equation with the velocity equation?

F(t)= (24.00kg*m/s^3) t - (28kg*m/s^2)

v(t)= (6.00m/s^3) t^2 - 14.00m/s (t) + 7.00m/s

Not quite sure how to multiply those two though...Help?

[B]d)How much work is done on the object for the time interval given?[/B]

Work= Integral F dx=

F(t)= (24.00kg*m/s^3 )t - (28.00m/s^2)

W= Integral F dx= integral(24.00kg*m/s^3 )t - (28.00m/s^2)=

(12.00kg*m/s^3)t^2 - (28.00m/s^2)t + 7.00m/s

this however is the same eqzn I got for power vs time eqzn


[B]I need the most help on part C[/B]


Thanks

Answer 1

hm..I guess I can't bold..

I thought bold was

[b] b[/b]

Answer 2

Bold is < b >

< / b >

Take the spaces out.

Answer 3

Oh..whoops I forgot..Thanks Ms Sue
unfortunately my post looks highly complicated and hard to read without the bold...

Answer 4

I'm sorry I can't help you with chemistry, Christina. My only chem class was in high school over 50 years ago! <g>

Answer 5

Thanks =D but I currently don't have any chemistry questions but rather a complicated <b>physics</b> question

Answer 6

(a) x(t)= 2t^3 -7.00 t^2 + 7.00 t
V(t) = 6t^2 -14 t +7
When t=0, V = 7 m/s
When t=2s, V = 24 -28 +7 = 3 m/s
Change in momentum = 2 kg*(-4 m/s)
= -8 kg m/s
(b) Plot m*(6t^2 - 14 t)
That will be the change in momentum vs. time.
(c) Power = Force * V = [(dV/dt)/M]*V
(d) For the work done on the object in a specific interval, compute the change in (1/2)M V^2 during the ionterval
= [(12t- 14)/2]*(6t^2 -14t +7)
= (6t-7)(6t^2 -14t +7)