1. ## winch raising a load

A winch is used to raise a 200 kg load. The maximum power of the winch is 5 kW. Calculate the greatest possible acceleration of the load when its speed is 2 ms^1, and the greatest speed at which the load can be raised.

Alright so this is how I went about things: power=force x velocity, so upward force = 5000/2 = 2500. So net upward force = 500 N, so greatest possible acceleration = 500/200 = 2.5 ms^2.
I know this is incorrect, where did I go wrong?

2. ## Re: winch raising a load

Originally Posted by furor celtica
A winch is used to raise a 200 kg load. The maximum power of the winch is 5 kW. Calculate the greatest possible acceleration of the load when its speed is 2 ms^1, and the greatest speed at which the load can be raised.

Alright so this is how I went about things: power=force x velocity, so upward force = 5000/2 = 2500. So net upward force = 500 N, so greatest possible acceleration = 500/200 = 2.5 ms^2.
I know this is incorrect, where did I go wrong?
$P = F \cdot v = ma \cdot v$

$a = \frac{P}{mv} = \frac{5000}{200(2)} = 12.5 \, m/s^2$

3. ## Re: winch raising a load

thanks for your time bud, but the correct answer is 0.2 ms^2

4. ## Re: winch raising a load

Originally Posted by furor celtica
thanks for your time bud, but the correct answer is 0.2 ms^2
yes, I forgot to take the work done by the force of gravity ... but taking that into account, I arrive at 2.5 m/s^2 also.

5. ## Re: winch raising a load

well that doesnt help me then does it

6. ## Re: winch raising a load

alrighty now i've got an issue with ths second problem, finding the greatest speed at which the load can be raised
so greatest speed means the upwards force is equal to the downwards force, so the upwards force will be 2000N
power being 5000, speed is 2.5 ms^1
however the answer is 2.55, another typo or did i truly get it wrong this time?

7. ## Re: winch raising a load

Originally Posted by furor celtica
A winch is used to raise a 200 kg load. The maximum power of the winch is 5 kW. Calculate the greatest possible acceleration of the load when its speed is 2 ms^1, and the greatest speed at which the load can be raised.

Alright so this is how I went about things: power=force x velocity, so upward force = 5000/2 = 2500. So net upward force = 500 N, so greatest possible acceleration = 500/200 = 2.5 ms^2.
I know this is incorrect, where did I go wrong?
$f= mg+m\dot{v}$

$\text{power}=\text{rate of doing work}=fv= (mg+m\dot{v})v$

which to me looks like $\dot{v}=2.5 \text{ m/s}^2$.

Consider the possibility that the known correct answer is wrong?

CB

8. ## Re: winch raising a load

Originally Posted by furor celtica
alrighty now i've got an issue with ths second problem, finding the greatest speed at which the load can be raised
so greatest speed means the upwards force is equal to the downwards force, so the upwards force will be 2000N
power being 5000, speed is 2.5 ms^1
however the answer is 2.55, another typo or did i truly get it wrong this time?
I agree with you, you set $\dot{v}=0$ and solve for $v$ that gives $2.5 \text{ m/s}$

CB