physics motion problem

[pitbull]

problem reads:
a ball of mass 0.500kg is attached to the end of a cord 1.50m long, the ball is whirled horizontaly on a friction less table, if the cord can withstand mass of 10.0kg a) what's speed of the ball.

i used Fr = mar = m(v^2/r)

i solved T max = m(v^2/r) *T max -

v = square root Tr/m
= (10.0*1.50m)/0.500kg
= 5.6m/s

now this is the max speed the ball can have before the string breaks, however they asked for the speed not the max it can take before the string breaks ??? are these two the same speed = max speed if not can someone show me.


another question:

a 1000kg car accelerates uniformly to speed of 30m/s in it's first 10s find a)work done on car in this time b)avg power delivered c)instantaneous power delivered by engine at t = 2s.

W= Kf-Ki
W= 1/2mvf^2- 1/2mvi^2 = 1/2m(vf^2-vi^2)

a) 450000 Jules

b)P = W/T = 450000/10s = 45000 Watts

c) instantaneous ???? how?

[stuffy]

For the first question, it is the same.

For the second questions, instantaneus power is the derivative of work or just the cross product between force and velocity.

[arcnets]

Hi pitbull,
1)
when dealing with min/max problems, it can be useful to use inequalities:

mv2/r <= Tmax
v <= [squ](Tmaxr/m)

Answer: The speed is equal to or smaller than ...

2)
I think instantaneous means dW/dT (the derivative).

[pitbull]

thanks for the response.

so dw/dt would be

4500000/2 = 225000 ?

or is it:

a= vt

= (30m/s)/10s
= 3 m/s

v= at
P= F*v
F= m*a
P= ma(at)
= ma^2*t
(1000)*(3^2)*2
= 18000 W

by the way are part a & b on the second question right?

[HallsofIvy]

Since part (a) does not give angular velocity, it probably means maximum possible velocity- but it's badly stated.

In particular "the cord can withstand mass of 10.0kg " makes no sense to me. What does it mean to say that a cord can withstand a mass?

I would guess that they mean the cord could support the weight of a 10 kg mass- that it's maximum tension would be 98 Newtons.

For the second problem, you will need to find the work done at any instant. The problem tells you that the acceleration is uniform so at any time t (in seconds from the time you begin the acceleration) the speed is (1/2)at2 where a is the acceleration. Since the vehicle attains 30 m/s in 10 s, (1/2)a(100)= 50a= 30 so a= 3/5= 0.6 m/s2 and the velocity at any instant (from t=0 to t= 10) is 0.3 t2. Use that to determine the work done at time t and differentiate to find the instantaneous power.