Recent content by francisco300

Homework Statement Basically, I need to calculate the mass of the pulley and friction using data that I collected. Homework Equations I was given: (m1 - m2)g = (m1 + m2 + mp)a + f The Attempt at a Solution For the first part, I collected data using a sensor to find the...

No, one of my classmates got the same answer so i kind of assumed. The question says work done by external forces on the lawn mower. Are the forces I have on my diagram external forces? If so, then what are the internal forces? Also, is net work the sum of all work no matter if positive or...

I got 200.2 Joules Which would me Work done by force which is= 50*cos 60 * 10m = 250 Minus 49.8 J from kinetic friction.

Oh I see now. I forgot to include the *293.3 in my Work equation but I got the answer correct. Thanks for the help

Thanks for the welcome! Yeah I can manage the rest of the forces, I just needed help with the work done by frictional force.

I sketched my free body diagram in paint. Does it look correct?

Well I know kinetic friction= μ*N Normal force≠ Weight in this case because the 50 N at 60° adds a downward force. So to calculate force I would use this equation: N-W-Fsin60= 0 I get Normal force = 293.3 N (I use 10 m/s^2 for gravity) So then Work for Kinetic friction = .017 *...

Quadrant 4 would be the bottom right quadrant in a graph. That is true but usually I like to label forces starting the tail from the center of the object. Is that incorrect? I am not sure what you mean by the last question? I believe because each force can do work depending on the angle...

Work Problem -- Pushing a lawnmower 1. A lawn mower (25kg) is pushed a horizontal distance of 10m by a 50 N downward force directed 60 degrees to the horizontal. The coefficient of kinetic frictional force is .017. What is the work done by each of the external forces on the lawn mower...