A block of weight w = 30.0 N sits on a frictionless inclined plane, which makes

Question

A block of weight w = 30.0 N sits on a frictionless inclined plane, which makes an angle = 35.0 with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 17.2 N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

PART A

What is Wg, the work done on the block by the force of gravity w  as the block moves a distance L = 4.50 m up the incline?

PART B

What is WF, the work done on the block by the applied force F  as the block moves a distance L= 4.50 m up the incline?

PART C

What is WN, the work done on the block by the normal force as the block moves a distance L = 4.50 m up the inclined plane?

This is the second time ive asked this question. The first time I recived incorrect answers so please be careful :).

Explanation / Answer

part(A)

W F*s*costheta

theta = anfle between F*s

the component of weight along the plane = W*sin35 down the plane

the distance moved = L = 4.5 m up the plane

theta = 180 degrees

Wg = W*sin45*L*cos180 = -30*sin35*4.5 = -77.43 J   <<<-------answer

part(B)

F & L are up the plane

theta = 0 degrees

WF = F*L = 17.2*4.5 = 77.4 N   <<<-------answer

part(c)

normal force = N = W*cos35 perpendicular to the incline surface

theta = 90

Wn = N*L*cos90 = 0 <<<-------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.