# Physics motion problem

pitbull
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?

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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.

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?

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