Physics...seriously seriously need help esp part C
posted by ~christina~ .
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(vfvi)= 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

hm..I guess I can't bold..
I thought bold was
[b] b[/b] 
Bold is < b >
< / b >
Take the spaces out. 
Oh..whoops I forgot..Thanks Ms Sue
unfortunately my post looks highly complicated and hard to read without the bold... 
I'm sorry I can't help you with chemistry, Christina. My only chem class was in high school over 50 years ago! <g>

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

(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)
= (6t7)(6t^2 14t +7)
Respond to this Question
Similar Questions

Physics
A woman on a bridge 90.0m high sees a raft floating at a constant sped on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 6.00m more to travel before passing under … 
physics
A novice golfer on the green takes three strokes to sink the ball. The successive displacements are 4.00m north, 2.00m northeast, and 1.00m at 30.0 degrees west of south starting at the same initial point, an golfer could make the … 
chemistry
A volume of 100mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence … 
physics
Hello everyone, I need help with this question: A dart gun is fired while being held horizontally at a height of 1.00m above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance … 
physics  relative motion
A ferryboat, whose speed in still water is 4.00m/s, must cross a river whose current is 3.00m/s. The river runs from west to east and is 128m wide. The boat is pointed north. a) If the boat does not compensate for the flow of the river … 
Chemistry
Calculate the pH of a solution that is 2.00M HF, 1.00M NaOH, and 0.393M NaG (Ka=7.2 x10^4). The answer I got was 4.76 .. 
Chemistry
A student determined the H of neutralization of nitric acid (HN03) mixed with NaOH solution, using the procedure described in this experiment. Fifty milliliters of 1.00M HN03 was added to 50.5 mL of 1.00M NaOH solution, and the following … 
physics 11
a 4.00kg object traveling westward at 25.0m/s hits a 15.0kg object at rest. the 4.00kg object bounces eastward at 8.00m/s. what is the speed and direction of the 15.0kg object? 
Physics
There are 2 objects connected with ideal, massless string that is hanging over an ideal (massless and frictionless) pulley. The angled surface is at a 25.0 degree angle. All surfaces are firctionless. A force F1 is applied horizontally … 
Physics
An interacting system of two objects has no external force acting on the system. The first has a mass of 2.00kg and is moving with a velocity (2.00m/s)i + (1.00m/s)j at time t=0s. The second object has a mass of 3.00kg and is moving …