# Oxford PAT 2014, Question 15

## 7 thoughts on “Oxford PAT 2014, Question 15”

1. Jaime says:

Hi,

Why isn’t the force that the motor is pulling up with mg? I’ve labelled the diagram and the tension in each string is mg/3; shouldn’t F = 3T, i.e. mg?

Thanks

2. Morte says:

Hey I have a doubt, at the first part the power of the motor wouldn’t it be giving kinetic energy and gpe to the mass? But kinetic energy is constant here so it doesn’t appear in our equation?

1. Your second question is spot on. In the situation described in the question, the mass is moving with constant upwards velocity of u, so the mass is neither losing nor gaining kinetic energy. So the motor is only transferring gpe to the mass, hence the equation VI = mug.

3. Harry says:

could you please explain why is the velocity of the string and wheel 3u and what equation did u use for angular velocity? thanks

1. Hi Harry, thanks for the question.

Why is the velocity of the string and the wheel 3u? Imagine the mass moves up by a small distance x. There are three lengths of string holding up the mass, and each of them will get shorter by an amount x, so the total shortening of the three lengths of string is 3x. Of course, the three lengths of string are actually made by looping one single length round two pulleys, so for every distance x that the mass moves up, the string has to move 3x over the top pulley. So if the mass moves at velocity u the string must be moving over the top pulley with velocity 3u.

What equation did you use for angular velocity? The velocity at the rim is the angular velocity times the radius of the wheel, so the angular velocity is the velocity at the rim divided by the radius. The radius is D/2, so the angular velocity is (3u)(2/D).

1. Harry says:

thanks