Solving Energy Questions: Force, Gravity, Kinetic Energy

  • Thread starter Thread starter kfink85
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion focuses on solving physics problems related to work and kinetic energy. For the first problem, the work done by the applied force is calculated as approximately 31.95 J, while the normal force and gravitational force do not contribute to work since they act perpendicular to the displacement. The net force on the block remains unclear to the poster, indicating a need for further clarification. In the second problem, the kinetic energy of the baseball at its highest point is questioned, with the poster unsure if they should consider vector components. The key takeaway is that only the horizontal component of velocity affects kinetic energy at the peak of the projectile's motion.
kfink85
Messages
12
Reaction score
0
Here are a couple problems that I have been attempting to do but I am not sure if I am doing them right.

1. A block of mass 2.50 kg is pushed 2.20 m along a frictionless horizontal table by a constant 16.0-N force directed 25 degrees below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gavity, and (d) the net force on the block.

(a) W = Fcos(pheta)*Delta X
W = ( 16 cos 25 )( 2.2 )
W= 31.95 J

(b) 0 since it is perpendicular to the displacement
(c) 0 since it is perpendicular to the displacement
(d) I'm not sure how to do this, someone please help :confused:

2. An outfielder throws a 0.150-kg baseball at a speed of 40.0 m/s and an initial angle of 30 degrees. What is the kinetic energy of the ball at the highest point of its motion?

I know that Vy= 0 m/s at the top of the motion but I'm not sure if this information is even relevant. Do I need to break this down into vectors or no?

I know KE=1/2mv^2. Do I just plug it in or is the velocity going to be changed at the top due to vectors?

Thank you in advance to anyone that can help.
 
Physics news on Phys.org
Anyone have any idea?
 
The Work done by the "net Force" is what changes the KE.

The horizontal component of velocity (at throw) is the only component at peak.
 
1 d) same as part a) because table is horizontal and firctinless, so only work acting on block is force of the push
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top