# Signs of work done and delta K.E.

• Ebby
In summary: W = |\vec F| \, |\vec s| \, \sin \theta = $$Again, the answers are negative regardless of the coordinate system. Ebby Homework Statement Find the work done by the braking force and the change in kinetic energy of the car. Relevant Equations F = m . a s = v_final^2 - v_initial^2 / 2 . a W = F . s delta K.E. = K.E._final - K.E._initial I'm asking about this with particular reference to the signs of the answers. Here is the question: The answers in the back of the book are: (a) ##1.2 \times 10^4 \text{ N}## (b) ##39 \text{ m}## (c) ##4.7 \times 10^5 \text{ J}## (d) ##4.7 \times 10^5 \text{ J}## Here's a rough sketch of the situation. I suppose it's not really necessary, but I feel the vectors do emphasise the idea of direction and sign. I'll now go through each part, with particular attention to the signs of the answers. I have some disagreements with the book, especially regarding parts (c) and (d). (a) The acceleration ##\vec a## is the negative ##x## direction, so the component ##F_x## must also be negative:$$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}$$Oops I accidentally pressed post thread. This isn't finished quite yet. How do I delete it? Last edited: Ebby said: Homework Statement: Find the work done by the braking force and the change in kinetic energy of the car. Relevant Equations: F = m . a s = v_final^2 - v_initial^2 / 2 . a W = F . s delta K.E. = K.E._final - K.E._initial I'm asking about this with particular reference to the signs of the answers. Here is the question: View attachment 328218 View attachment 328219 The answers in the back of the book are: (a) ##1.2 \times 10^4 \text{ N}## (b) ##39 \text{ m}## (c) ##4.7 \times 10^5 \text{ J}## (d) ##4.7 \times 10^5 \text{ J}## Here's a rough sketch of the situation. I suppose it's not really necessary, but I feel the vectors do emphasise the idea of direction and sign. View attachment 328283 I'll now go through each part, with particular attention to the signs of the answers. I have some disagreements with the book, especially regarding parts (c) and (d). (a) The acceleration ##\vec a## is the negative ##x## direction, so the component ##F_x## must also be negative:$$F_x = m \cdot a_x = 1500 \cdot -8 = -1.2 \times 10^4 \text { N}$$The book has this answer as being positive. I guess what they're really asking for is the magnitude of the force. OK, accepted. That's what braking force means. The sign is only relevant once you have chosen a coordinate system. You could have the car moving in the negative ##x## direction (or any other direction). Ebby said: (b) No problems here. ##|\vec s|## and ##s_x## happen to be the same.$$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.

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

## What is the relationship between work done and kinetic energy?

The relationship between work done and kinetic energy is described by the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Mathematically, W = ΔK.E., where W is the work done and ΔK.E. is the change in kinetic energy.

## How do you calculate work done on an object?

Work done on an object is calculated using the formula W = F * d * cos(θ), where W is the work, F is the force applied, d is the displacement of the object, and θ is the angle between the force and the direction of displacement. If the force is in the direction of displacement, cos(θ) is 1.

## What are the units of work done and kinetic energy?

The units of work done and kinetic energy are both joules (J) in the International System of Units (SI). One joule is equivalent to one newton meter (N·m), where a newton is the unit of force and a meter is the unit of displacement.

## Can work done be negative, and what does that signify?

Yes, work done can be negative. Negative work occurs when the force applied to an object is in the opposite direction of its displacement. This typically signifies that the object is losing kinetic energy, for example, when friction or another opposing force is acting on it.

## What is kinetic energy and how is it calculated?

Kinetic energy is the energy that an object possesses due to its motion. It is calculated using the formula K.E. = 0.5 * m * v^2, where K.E. is the kinetic energy, m is the mass of the object, and v is its velocity. This formula shows that kinetic energy is directly proportional to the mass of the object and the square of its velocity.

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