# Determine the work done by the applied force

1. Oct 14, 2006

### BunDa4Th

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.

2. Oct 14, 2006

### gunblaze

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. Oct 15, 2006

### BunDa4Th

that is all the info was given and thats the exact question given.

I think the force is acting 25 degrees downwards from the block.

4. Oct 15, 2006

### gunblaze

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 travelled which i think u got it. For both b and c, the WD equals zero because there is no distance travelled 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. Oct 15, 2006

### BunDa4Th

Thanks, that make much more sense now.