3. To push a 25-kg crate up a 27 incline, a virker force of 120 N, parallel to the incline. As up. 3.6 m, how much work is done on the TD worker, (b) the force of gravity, and (c) them the incline?
Answers
Answered by
0
Answer:
Wc = mg = 25kg * 9.8N/kg = 245N. = Weight of crate.
Fc = 245N @ 27deg.
Fp = 245sin27 = 111.2N.=Force parallel to incline.
Fv = 245cos27 = 218.3N. = Force perpendicular to incline = Normal.
a. W = Fd = 120 * 3.6 = 432J.
b. W = Fc*d=Fc*h=245 * 3.6sin27 = 400J.
c. W=Fv * d = 218.3 * 3.6sin27 = 357J.
Answered by
1
Answer:
Good morning sir today working plan Bahadurgarh
Good morning sir today working plan Bahadurgarh
Similar questions