Determine the work done by the applied force

In summary, the conversation discusses a problem involving a block being pushed along a frictionless table by a constant force. The work done by the applied force, normal force, gravity, and net force on the block are calculated. There is confusion about the direction of the force and the work done by the normal and gravitational forces. The direction of the force is clarified to be 25 degrees below the horizontal and the calculations for the work done are confirmed to be correct.
  • #1
BunDa4Th
188
0
Okay I am confuse on how i came to this conclusion and how i came to that answer.

A block of mass 2.50 kg is pushed 2.80 m along a frictionless horizontal table by a constant 12.0 N force directed 25.0° below the horizontal.

(a) Determine the work done by the applied force.
30.45J
(b) Determine the work done by the normal force exerted by the table.
0 J
(c) Determine the work done by the force of gravity.
0J
(d) Determine the work done by the net force on the block.
30.45J

I am confuse on b, c, d. I know that to determine the applied force i would use W = (Fcostheta)(deltaX).

For b i would use W = (mgcos90)deltaX is that correct? because since the block is on a table which is flat and normal force is pointing up so from table to normal force it makes a 90 degree angle.

For c I did a guess since I was not sure but i thought that it would be zero since there is no incline and gravity is not acting on this block because its not falling.

D i came up with Wnet = W_F + W_f + W_N + W_mg
W_F = 30.45
W_f = 0 (no friction so it is zero)
W_N = 0 (not sure why)
W_mg = 0 (not sure why)

sorry its a bit long but can anyone explain this to me.
 
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  • #2
can u clarify the direction of the force on the object? what u mean by a constant force of 12 N directed below the horizontal? So the force is acting 25 degrees upwards on the block or downwards?
 
  • #3
that is all the info was given and that's the exact question given.

I think the force is acting 25 degrees downwards from the block.
 
  • #4
For a, b and c, the answers u provided are right.. if the force on the block is really acting downwards from the block. for a) All u have to do is to take the horizontal component of the force on the block and multiply it by the distance traveled which i think u got it. For both b and c, the WD equals zero because there is no distance traveled in the direction of the force. for d, all u got to do is to sum up the WD on the block since they are scalars which will actually give u the wd done by moving the block, 30.45.
 
  • #5
Thanks, that make much more sense now.
 

Related to Determine the work done by the applied force

1. What is the definition of work in physics?

The definition of work in physics is the product of the force applied on an object and the distance the object moves in the direction of the force.

2. How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. The formula for work is W = F * d, where W is work, F is force, and d is distance.

3. What is the unit of work in physics?

The unit of work in physics is the joule (J). One joule is equal to one newton-meter (N*m). Other common units of work include kilojoules (kJ) and calories (cal).

4. How is the direction of work determined?

The direction of work is determined by the direction of the force applied on an object. If the force is applied in the same direction as the object's motion, the work is considered positive. If the force is applied in the opposite direction of the object's motion, the work is considered negative.

5. Can work be negative?

Yes, work can be negative. This occurs when the force applied on an object is in the opposite direction of the object's motion. Negative work indicates that energy is being transferred out of the system, while positive work indicates that energy is being transferred into the system.

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