Signs of work done and delta K.E.

[Ebby]

Homework Statement

Find the work done by the braking force and the change in kinetic energy of the car.

Relevant Equations

F = m . a
s = v_final^2 - v_initial^2 / 2 . a
W = F . s
delta K.E. = K.E._final - K.E._initial

I'm asking about this with particular reference to the signs of the answers. Here is the question:

The answers in the back of the book are:

(a) $1.2 \times 10^4 \text{ N}$
(b) $39 \text{ m}$
(c) $4.7 \times 10^5 \text{ J}$
(d) $4.7 \times 10^5 \text{ J}$

Here's a rough sketch of the situation. I suppose it's not really necessary, but I feel the vectors do emphasise the idea of direction and sign.

I'll now go through each part, with particular attention to the signs of the answers. I have some disagreements with the book, especially regarding parts (c) and (d).

(a) The acceleration $\vec a$ is the negative $x$ direction, so the component $F_x$ must also be negative:$$F_x = m \cdot a_x = 1500 \cdot -8 = -1.2 \times 10^4 \text { N}$$The book has this answer as being positive. I guess what they're really asking for is the magnitude of the force. OK, accepted.

(b) No problems here. $|\vec s|$ and $s_x$ happen to be the same.$$s_x = \frac {{v_x}_f^2 - {v_x}_i^2} {2 \cdot a_x} = \frac {0 - \left( \frac {90} {3.6} \right)^2} {2 \cdot -8} = \frac {-625} {-16} = 39 \text { m}$$

Oops I accidentally pressed post thread. This isn't finished quite yet. How do I delete it?

 

[PeroK]

Homework Statement: Find the work done by the braking force and the change in kinetic energy of the car.
Relevant Equations: F = m . a
s = v_final^2 - v_initial^2 / 2 . a
W = F . s
delta K.E. = K.E._final - K.E._initial

I'm asking about this with particular reference to the signs of the answers. Here is the question:

View attachment 328218
View attachment 328219

The answers in the back of the book are:

(a) $1.2 \times 10^4 \text{ N}$
(b) $39 \text{ m}$
(c) $4.7 \times 10^5 \text{ J}$
(d) $4.7 \times 10^5 \text{ J}$

Here's a rough sketch of the situation. I suppose it's not really necessary, but I feel the vectors do emphasise the idea of direction and sign.
View attachment 328283

I'll now go through each part, with particular attention to the signs of the answers. I have some disagreements with the book, especially regarding parts (c) and (d).

(a) The acceleration $\vec a$ is the negative $x$ direction, so the component $F_x$ must also be negative:$$F_x = m \cdot a_x = 1500 \cdot -8 = -1.2 \times 10^4 \text { N}$$The book has this answer as being positive. I guess what they're really asking for is the magnitude of the force. OK, accepted.

That's what braking force means. The sign is only relevant once you have chosen a coordinate system. You could have the car moving in the negative $x$ direction (or any other direction).

(b) No problems here. $|\vec s|$ and $s_x$ happen to be the same.$$s_x = \frac {{v_x}_f^2 - {v_x}_i^2} {2 \cdot a_x} = \frac {0 - \left( \frac {90} {3.6} \right)^2} {2 \cdot -8} = \frac {-625} {-16} = 39 \text { m}$$

(c) $$W = |\vec F| \, |\vec s| \, \cos \theta = $$

If you were going to say that the answers to parts c) and d) should be negative, then I agree with you. These are negative values regardless of the coordinate system.

 

[Ebby]

Yes that's exactly what I was going to say! Thanks :)