A 3.9-kg block is pushed 3.1 m up a vertical wall with constant speed by a const

Question

A 3.9-kg block is pushed 3.1 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 the figure below. If the coefficient of kinetic friction between block and wall is 0.30, determine the following.

(a) the work done by

J

(b) the work done by the force of gravity
J

(c) the work done by the normal force between block and wall
J

(d) By how much does the gravitational potential energy increase during the block's motion?
J

Explanation / Answer

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 = 3.9 × 9.8 = 38.22
F = 38.22 / (0.5 0.3 × 0.866025) = 159.12 N.

a) W = F r cos , where 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 = 159.12 × 4 cos 60° = 246.636 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 . Now, actually F means weight, mg and = 180°; so W = mgr cos 180° = 3.9 × 9.81 × 4 = 153.036 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 , where = 90°; thus, W = N r cos 90° = 0 J, since cos 90° = 0.

d) U = mgh = 3.9 × 9.8 ×3.1 = 118.482 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 246.636 J, and only 118.482 J can be recovered, then 128.15 J is the energy loss due to friction.

Check: work done against friction: W = f r cos = µN r cos 180° = 0.3 × 159.12 cos 30° × 3.1 = 128.15J

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

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