Help with a couple physics problems

[masteroftheuniverse]

I am having some trouble with a couple of physics h/w problems and would appreciate any help you can give me..

Julie pulls a 61.5 N bag a distance of 278 m at a constant velocity. The force she exerts is 48.4 N at an angle of 66.8 degrees above the horizontal.
I found the work julie does on the bag to be 5300 J.
Yet I am not sure how to find A) the work done by the force of friction on the bag and B) the coefficient of kinetic friction on the bag

Next problem is
A rain cloud contains 2.82 x 10^7 kg of water vapor.
The acceleration of gravity is 9.81m/s^2
How long would it take a 2.10 kW pump to raise the same amount of water to the cloud's altitude of 1.44km? Answer in seconds (s)
[I tried Power= work(FsCOS[theta]) divided by t. So 2100w=2.82x10^7(9.81)1440m COS 90 degrees DIVIDED by t, which just equals 0 since cos90 is 0. I DONT GET IT!]

 

[cesiumfrog]

why is theta 90? why is this not in the Homework forum?

 

[kyle8921]

Sure, I'd be happy to help you with these physics problems.

For the first problem, we can start by finding the work done by Julie on the bag using the equation W = Fdcosθ, where W is work, F is the force, d is the distance, and θ is the angle between the force and the displacement. Plugging in the given values, we get:

W = (48.4 N)(278 m)cos(66.8°) = 5300 J

This is the work done by Julie on the bag. To find the work done by friction, we can use the fact that the total work done on an object is equal to its change in kinetic energy. Since the bag is moving at a constant velocity, its kinetic energy does not change, so the work done by friction must be equal to the negative of the work done by Julie. Therefore, the work done by friction is -5300 J.

To find the coefficient of kinetic friction, we can use the equation W = μN, where μ is the coefficient of kinetic friction and N is the normal force. In this case, N is equal to the weight of the bag, which is 61.5 N. Plugging in the values, we get:

-5300 J = μ(61.5 N)

Solving for μ, we get μ = -5300 J / 61.5 N = -86.2. This is not a physically meaningful value, so we can conclude that there must be an error in the given information or in our calculations.

Moving on to the second problem, we can use the equation P = W/t to find the time it takes for the pump to raise the water to the cloud's altitude. However, there are a few issues with the values given in this problem. First, the power of the pump is given in watts, but the work done by the pump is measured in joules, so we cannot use the given power value directly. We need to convert it to joules per second (J/s) by dividing by 1000. Second, the given value for the mass of water (2.82 x 10^7 kg) is not used in the given equation, so it must be a mistake. Instead, we can use the mass of water (in kilograms) multiplied by the acceleration due to gravity (in m/s^2) to find the weight of the water,