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P= F.v, where P is power, and F.v is the dot product of the Force and velocity vectors.
Yes, that looks correct, if you use only the magnitudes of the vectors, i.e. the "dot" product.Force x speed = force x (distance / time) = (force x distance) / time = work (energy) / time = power.
P=v/f=======================Well, let's see.Force = newtonWork = force x distance = newton-meterPower = work/time = (newton-meter)/(second)The first answer above says Power = (Velocity)/(force). Let's check it out.Speed= length/time = meter/secondForce = newtonV/f = (meter)/(newton-second)Power = (newton-meter)/(second)We don't know what kind of monstrosity V/f is, but it's not power.How about (force) x (speed) = (newton) x (meter/second) = (newton-meter)/(second).That's a lot nicer.In all its glory, Power = F · V .You, the questioner, said 'velocity', not 'speed', so you're going to have toface this for what it is ... the product of two vectors. Force is a vector, andso is velocity, but power is not. The way you multiply two vectors and geta scalar is by means of the 'vector dot-product'. The dot-product of forceand velocity is:(magnitude of the force) times (magnitude of the velocity) times (cosine of the angle between them).The reason for this is: If the force isn't pushing in the same direction as the velocity,then not all of it produces power, only the part of it that points in the right direction.If you're trying to push a heavy wagon and it's not moving fast enough, whatdo you do ? You let your feet get farther behind the wagon, and you crouchdown so that your shoulders are lower and more in line with the wagon. Thereason you do that is: Only the part of the force that lines up with the motionhelps with the motion, so if you want to push faster, you get the force down towhere it lines up better with the motion. You reduce the angle between theforce and the motion. That makes the cosine of the angle greater, so the dotproduct is greater. Even though the magnitude of the force hasn't changed,the component of the force in the direction where you need it has becomegreater, by reducing the angle between them.
Work, in physics, is defined as the product of force x distance. This assumes that the force is constant, and that it is in the same direction as the movement. Otherwise, a slightly more complicated formula is used: integral of (force dot-product ds), where ds is a short amount of movement.
Power is the time derivative of energy, E. Energy can be scalar or vector. Thus power can be scalar or vector. Energy is a quaternion and consists of a scalar or real part Er and a vector part Ev. Energy E=Er + Ev, for example E= FR = -F.R + FxR = -FRCos(x) + FRsin(x). The real part is a scalar called "Energy" and the vector part is called "Torque" but has the same units Joules. Energy is defined by the units. P=dE/dt = d(Er + Ev)/dt = dEr/dt + dEv/dt = Pr + Pv. Power can be a scalar or a vector or both.
P= F.v, where P is power, and F.v is the dot product of the Force and velocity vectors.
There are two likely calculus applications of this problem. Both differential calculus and basic vector operations can be used to solve for power in a scenario, depending on how a problem is defined. Power is the dot-product of a force vector and a velocity vector and... Power is a change in energy over time, or in differential terms: dE/dt If you were given a function that defined a system's energy with respect to time, you could derive it to find a function for that system's power output. If you were given a force vector and a velocity vector and asked to find the total power applied to the system, you could take the dot product of the two vectors to find this. Or, if you are not taking a calculus approach to it: Average power is simply energy divided by time The magnitude of power given a force and velocity can be found with the formula: P=F*v*cos(theta) Where F is the magnitude of the force v is the magnitude of the velocity theta is the angle between the two quantities.
Yes, that looks correct, if you use only the magnitudes of the vectors, i.e. the "dot" product.Force x speed = force x (distance / time) = (force x distance) / time = work (energy) / time = power.
Power is force times velocity times the cosine of the angle between the force vector and the velocity vector (i.e the "dot product", "scalar product", or "internal product" of the force and velocity vectors). We are not told the angle between the two vectors in this case, but if we assume that both are measured in the same direction, the angle between them is zero and its cosine is one.P = F•v= |F|*|v| * cos(θ)= 700 N * 6m / 6 s * cos 0°= 700 WThe power used is 700 Watts.Power for the muscles is provided by organelles called "mitochondria". These must not be confused with midichlorians, nor with chloroplasts, for that matter.
Yes and no. It's the dot product, but not the cross product.
As we know work done is a scalar. Also the work done is referred to as the product of force and displacement. so, we consider the dot product of force and displacement which would result in a scalar.
P=v/f=======================Well, let's see.Force = newtonWork = force x distance = newton-meterPower = work/time = (newton-meter)/(second)The first answer above says Power = (Velocity)/(force). Let's check it out.Speed= length/time = meter/secondForce = newtonV/f = (meter)/(newton-second)Power = (newton-meter)/(second)We don't know what kind of monstrosity V/f is, but it's not power.How about (force) x (speed) = (newton) x (meter/second) = (newton-meter)/(second).That's a lot nicer.In all its glory, Power = F · V .You, the questioner, said 'velocity', not 'speed', so you're going to have toface this for what it is ... the product of two vectors. Force is a vector, andso is velocity, but power is not. The way you multiply two vectors and geta scalar is by means of the 'vector dot-product'. The dot-product of forceand velocity is:(magnitude of the force) times (magnitude of the velocity) times (cosine of the angle between them).The reason for this is: If the force isn't pushing in the same direction as the velocity,then not all of it produces power, only the part of it that points in the right direction.If you're trying to push a heavy wagon and it's not moving fast enough, whatdo you do ? You let your feet get farther behind the wagon, and you crouchdown so that your shoulders are lower and more in line with the wagon. Thereason you do that is: Only the part of the force that lines up with the motionhelps with the motion, so if you want to push faster, you get the force down towhere it lines up better with the motion. You reduce the angle between theforce and the motion. That makes the cosine of the angle greater, so the dotproduct is greater. Even though the magnitude of the force hasn't changed,the component of the force in the direction where you need it has becomegreater, by reducing the angle between them.
Dot product and cross product are used in many cases in physics. Here are some examples:Work is sometimes defined as force times distance. However, if the force is not applied in the direction of the movement, the dot product should be used. Note that here - as well as in other cases where the dot product is used - the product is greatest when the angle is zero; also, the result is a scalar, not a vector.The cross product is used to define torque (distance from the axis of rotation, times force). In this case, the product is greatest when the two vectors are at right angles. Also - as in any cross product - the result is also a vector.Several interactions between electricity and magnetism are defined as cross products.
A Dot product is a very useful tool in both mechanics and 3D graphics. It calculates the cosine of the angle between two vectors.In two-dimensional space, the dot product of vectors [a, b] and [c, d] is ac + bd.Mechanical work is the dot product of force and displacement vectors.Magnetic flux is the dot product of the magnetic field and the area vectors.
it is the dot product of displacement and force . i.e. Fdcos(A) where F is the magnitude of force , d is the magnitude of displacement and A is the angle between them
Dot Products in Physics denote scalar results fmo vector products, e.g Work = F.D = FDCos(FD) a scalar result from the dot product of two vectors, F Force and D Displacement.
The dot-product of two vectors is the product of their magnitudes multiplied by the cosine of the angle between them. The dot-product is a scalar quantity.