Recent content by TheFlemster

so the correct answer would be a 831.02 N normal force acting on the person?

Fnet=N-mg Fnet=ma combining the two, ma=N-mg then, N=ma + mg N=(74)(1.43) + (74)(9.8) N=831.02

N=mg

The normal force and the force due to gravity?

F=mg + ma F=(74)(9.8) + (74)(1.43) F=831.02 N Wouldn't this be the total force acting on the person?

So then I ignore the acceleration of 1.43 m/s^2? I am still confused as to which way I am supposed to work it

Homework Statement A 6.50 kg block is placed on an inclined surface. The coefficients of static and kinetic friction between the block and the surface are 0.685 and 0.485, respectively. The incline makes an angle of 18.8° relative to horizontal. The block is at rest and remains at rest. What is...

Homework Statement A 74.0 kg person is standing inside an elevator. The elevator is moving from the 3rd floor to the 21st floor. As the elevator passes the 4th floor it is moving at 2.30 m/s and is increasing speed at a rate of 1.43 m/s2 . At this moment, what is the normal force that acts on...

Homework Statement A thin walled spherical shell with a mass of 0.605 kg and a radius of 0.0402 is released, from rest, at the top of an incline. The spherical shell rolls down the incline without slipping. The spherical shell takes 7.49 s to get to the bottom of the incline. A solid sphere...

Thanks to both of yall. I found the height with trig and recalculated and got the correct answer.

the first pic are the possible answer choices

Homework Statement A 5.72 kg block is released from rest on an inclined surface. The incline makes an angle of 25.0 degrees relative to horizontal. The block is known to be traveling at a speed of 1.33 m/s after sliding a distance of 2.60 m along the incline. How much work was done on the block...

Ok so i figured out all of number 1 and parts a,b,c, and d on number 2. Part e on number 2 is asking for the firctional force that acts on the block, i did 21.5N(cos(29.4)) = 18.7N. the answer is -18.7N. I know how to work it, but i do not understand why the frictional force is negative?

31.2N is the applied force given in the problem.