U don't have to solve this, can U just tell me which formulas to use? I'm not getting this long one at all...

Question

A block mass 2.50 KG is pushed 2.2 M along a frictionless horizontal table by a constant 16 N force directed 25degrees below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table (c) the force of gravity, and (d) the net force on the block.

= HOW TO FIND THEM ALL? ESPECIALLY A.

Answer 1

d) first: Net (unbalanced) force = 16Ncos 25
c) & b) : No work since there is no motion in these directions
a): W = f*d, so W = 16Ncos25 * 2.2m

There are some semantics at work here. If you take "applied force" to mean 16N@25°, not ALL that force does any work, just the horizontal component of it. Then again, if that force were absent, there wouldn't be any compnents of it to do ANY work........!

Answer 2

First of all let F = the applied force, N = the normal force on the mass from the table, and W = the weight (force due to gravity) on the object.

Do a free body diagram (can't help you with that here)

Work done by any force is the product of the component of the force acting in the direction of travel of the object times the distance travelled in that direction (yes that is a dot product).


(a)Unless the table gives way beneath the applied force and the weight of the object, the applied force does an amount of work equal to

F* d * cosine (theta), where theta is the angle the force F is directed at relative to the horizontal table.

(b) From your free body diagram, again, unless the table gives way, the Normal force exerted by the table is only in the vertical direction, upwards, and it is balanced by both tthe weight of the object and the component of the applied force F in the vertical direction, or,

N = W + F * sine(theta)

(c) You must do this before (b) [[don't you hate it when books ask you to answer something before you have the answer]]

W = mg, where m is the mass of the block and g = 9.8 m/s^2

(d) The net force on the block during the time the applied force is acting is the amount of the applied force in the horizontal direction, or

Fnet = F * cosine(theta)

****REMEMBER THAT FORCE ARE VECTORS AND THAT ALL ANSWERS ABOVE MUST HAVE A DIRECTION ASSOCIATED WITH THEM****

Answer 3

well its been a while since i took physics, but here it goes:
(wish i could draw this, it would make so much more sense)
A) work is force times distance W=F*D but since the force not along the same vector as the distance you have to multiply by the COS of the angle, ie. you have to breakdown the force into its horizontal and vertical vectors and only consider the horizontal in this case doing the work, so work=force * distance * COS (angle between them)
B) since the table is horizontal the normal force exerted by the table is in the vertical direction.
since the block is not moving in the vertical direction, the normal force is canceling all the downward forces,
so it is equal to sum of all the forces in the downward direction, gravity plus the vertical component of the external force (16*SIN 25)
C) force is mass times acceleration, in case of gravity mass of the object multiplied by the constant g (9.8 m/s^2)
D) you can draw a force diagram for the block, include the normal force vertically upward, gravity vertically downward, and 16N at 25 degrees below horizontal, which we already broke down to vertical (16*sin25) and horizontal (16*cos25)
notice the block is only moving in the horizontal direction, so you know the net force is in only horizontal (that is why in B the upward normal force had to cancel out the downward forces) so the only thing left is the horizontal component of the 16N force which is 16*cos25 (this is the only part of the force left to do any work)

Caution: am i missing anything? its been a while....

Answer 4

Work done is defined as F.s (dot product of force and displacement),

Thus W = F . s . cos (angle)

a) You can directly substitute

b) The displacement is along the table, the normal force is perpendicular and hence no work is done. (angle = 90, cos 90 = 0)

Answer 5

a) W = F*d*cos(angle between them) = [16 N *cos25] * 2.2 m
b) 2.5 kg * g + 16 N *sin25
c) 2.5 kg * g
d) 2.5 kg * g @ vertical + 16 N @ 25 degrees below the horizontal

Answer 6

beauty b You are on the big news!!…
http://www.osoq.com/funstuff/extra/extra04.asp?strName=beauty_b