A 5.0 kg block is pushed 4.0 m up a vertical wall with constant speed by a constant force of magnitude F?

Question

A 5.0 kg block is pushed 4.0 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of = 30° with the horizontal, as shown in Figure P5.70. If the coefficient of kinetic friction between block and wall is 0.30, determine the work done by each of the following.


(a) the force
J
(b) gravity
J
(c) the normal force between block and wall
J
(d) By how much does the gravitational potential energy increase during the block's motion?
J

thanks that works accept on my Hw it is telling me F is wrong? i did the math myself and its coming out the same.. help??

Answer 1

In order to answer a), you'll have to find F first. Since the block moves at constant velocity, motion is not accelerated. This means net force on block is zero. Consider vertical components of force. Pointing downward, there's weight and friction. Counteracting this, there's the vertical component of F, pointing upward. Thus,

F sin 30° = mg + µN.

But N = F cos 30°, so

F sin 30° = mg + 0.3 F cos 30°.

Solving for F,

F (sin 30° − 0.3 cos 30°) = mg = 5 × 9.8 = 49
F = 49 / (0.5 − 0.3 × 0.866025) = 204.00 N.

a) W = F r cos φ, where φ is the angle between vectors F and r. (Find this angle drawing F first; then, draw r, from the same starting point you used for F). You'll see that this angle is 60°. Then, F = 204 × 4 cos 60° = 408 J.

Another method: the vertical component of F is F sin 30°, as you can see from the figure in your book. Displacement is vertical, along the same direction as the vertical component of F. So, W = (F sin 30°) × 4. This yields the same figure as before, since sin 30° = cos 60°.

b) W = F r cos φ. Now, actually F means weight, mg and φ = 180°; so W = mgr cos 180° = −5 × 9.8 × 4 = 196 J. The negative sign means gravity doesn't do any work; instead, work is done against gravity.

c) Normal force N is just F cos 30°; W = N r cos φ, where φ = 90°; thus, W = N r cos 90° = 0 J, since cos 90° = 0.

d) U = mgh = 5 × 9.8 × 4 = 196 J. This is the same as the work done against gravity.

U represents the "recoverable" portion of total work done by force F. The remaining fraction is lost as heat, and is nor recoverable. Since total work done by F amounts to 408 J, and only 196 J can be recovered, then 212 J is the energy loss due to friction.

Check: work done against friction: W = f r cos φ = µN r cos 180° = −0.3 × 204 cos 30° × 4 = 212 J.

Ed. Odu's answers for the first two items are wrong. Vertical component of F is supposed not only to support the block, but do work against friction as well. His figure is too low; that wouldn't even support the block: 64.5 sin 30° = 32.25 N. Block's weight is 49 N.

Accordingly, his result for work done by F is lacking. His figure just accounts for potential energy increase. What about work done against friction?

Answer 2

First, let's find the magnitude of F. Since the speed is constant, the sum of forces is zero

m*g-F*cos(30)*.3=F*sin(30)

solve for F
F=(5*9.8)/(sin(30)+cos(30)*.3)
F=64.5 N

The work done by the force is
F*sin(30)*4+F8cos(30)*.3*4
=196 J

Gravity is
-196 J

Normal force is 0

PE increased by 196 J

Answer 3

consistent velocity Fcos(alpha) - ok(mg-Fsin(alpha)) = 0 F[cos(alpha)+ksin(alpha)] = kmg F= kmg/[cos(alpha)+ksin(alpha)] alphal = sixty seven.5 (no longer enable minus sign) F = paintings = sFcos(alpha) calculate your self