Recent content by jongro

Take this equation: f(x) = u0.5(2 - 2u) + (2u - u2)(0.5u-0.5) My lecturer simplified it to this: f(x) = (2u(2 - 2u) + (2u - u2)) / (2u0.5) = (6u - 5u2) / (2u0.5) My intuition tells me to simplify it like this: f(x) = 2u0.5 - 2u1.5 + u0.5 - 0.5u1.5 = sqrt(u)(3 - (5/2)u)...

Since friction is in the opposite direction as the direction of movement, wouldn't you expect the work to be negative? When I put it in the work formula, W = F * cos(180O) * d it comes out positive because F is negative and cos(180) is negative too. This means that if I have another force...

Sorry, I forgot the force: It's 20 Newtons from the dog's mouth (15 degrees form horizontal).

Homework Statement This is a question about work: There is a dog pulling a sled across a patch of snow *with a force of 20 Newtons*, the dog's mouth is 15 degrees above horizontal and the sled weighs 5Kg. The snow has a coefficient of friction of 0.2, how much work has the dog done once it has...