A train that had a mass of 18200 kg was powered across a level track by a jet engine that produced a thrust of 521000N for a distance of 444m. Find the work done on the train.
I thought this was pretty straightforward; however, it's apparently not.
I used W=Fd, but I got the wrong...
Ok, so gravitational force is 9.8*12, right? That would be 117.6N. So the downward force is 117.6+96.33228=213.93228. But that means ma does not equal the product of the two. Where am I messing up?
A 12kg block is released from rest on an inclined plane with angle 35. Acceleration of the block is 1.23457 The acceleration of gravity is 9.8m/2^2.
What is the coefficient of kinetic friction for the incline?
I found the normal force (12*9.8*cos35)=96.33228...
Ok, I found the frictional force by using Ff=mg*sin33...it gave me 90.73686 (the right answer-yay!). So, how do I find the max angle? I think I have to use tangent, but I don't know on what or where or anything!
A 17 kg block is at rest on an incline with angle 33. The coefficients of static and kinetic friction are 0.7 and0 0.59, respectively. The acceleration of gravity is 9.8m/s^2.
1) What is the frictional force actin on the 17kg mass?
2) What is the larges angle the incline can have so...
Mass of 2.6kg lies on a frictionless table, pulled by another mass of 4.1kg under the influence of gravity (mass 4.1kg is hanging off the edge of the table). The acceleration due to gravity is 9.8m/s^2. What is the magnitude of the net external force (gravitational) acting on the two masses...