Solving Energy Questions: Force, Gravity, Kinetic Energy

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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.
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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.
 
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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
 
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