<li><p>A person pushes a 14.5-kg lawn mower at constant speed with a force of 88.0 N directed along the handle, which is at an angle of 45.0 degrees to the horizontal. Calculate (a) the horizontal retarding force on the mower, then (b) the normal force exerted vertically upward on the mower by the ground, and (c) the force the person must exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds (assuming the same retarding force).</p></li>
<li><p>At the instant a race began, a 65-kg sprinter was found to exert a force of 800 N on the starting block at a 22 degrees angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? (b) If the force was exerted for 0.38 s, with what speed did the sprinter leave the starting block?</p></li>
<li><p>One 3.0-kg paint bucket is haning by a massless cord from another 3.0-kg paint bucket, also hanging by a massless cord. (a) If the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.60 m/s^2 by the upper cord, calculate the tension in each cord.</p></li>
<li><p>A 6500-kg helicopter accelerates upward at 0.60 m/s^2 while lifting a 1200-kg car. (a) What is the lift force exerted by the air on the rotors? (b) What is the tension in the cable (ignore its mass) that connects the car to the helicopter?</p></li>
<li><p>A window washer pulls herself upward using the bucket-pulley apparatus. (a) How hard must she pull downward to raise herself slowly at constant speed? (b) If she increases this force by 10 percent, what will her acceleration be? The mass of the person plus the bucket is 65 kg.</p></li>
</ol>
<p>Hmm... no one wants to offer some advice?</p>
<p>for q 28, draw FBD and find F in X dimension if im not wrong..</p>
<p>29a) draw triangle</p>
<p>F in the x direction = 800cos22 =741.75 N</p>
<p>Fx=ma (a in x direction)
741.75 N=65kg (a)
a=11.41 m/s squared</p>
<p>29b) V=Vinitial + at
V=0 + (11.41)(.38)
V= 4.34 m/s</p>
<p>31) m=helicoper+car
m=1200+6500 = 7700</p>
<p>Fg=mg
Fg=(7700)(9.8)
Fg=75460 N</p>
<p>31b) Fg=mg
Fg=1200 (9.8)
Fg=11760 N</p>
<p>thats all i got</p>
<p>Thanks man. That really helped me on some.</p>
<p>i have the same book...and we did this 3 weeks ago....but we only get odd problems so that we can check our answers at home</p>
<p>For number 31, the dude gave me the wrong answers. How do you do it so that the answer is the same as that of the book?</p>
<p>Does anyone know how I can do 28, 30, and 32, reproduced below?</p>
<ol>
<li><p>A person pushes a 14.5-kg lawn mower at constant speed with a force of 88.0 N directed along the handle, which is at an angle of 45.0 degrees to the horizontal. Calculate (a) the horizontal retarding force on the mower, then (b) the normal force exerted vertically upward on the mower by the ground, and (c) the force the person must exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds (assuming the same retarding force).</p></li>
<li><p>One 3.0-kg paint bucket is haning by a massless cord from another 3.0-kg paint bucket, also hanging by a massless cord. (a) If the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.60 m/s^2 by the upper cord, calculate the tension in each cord.</p></li>
<li><p>A window washer pulls herself upward using the bucket-pulley apparatus. (a) How hard must she pull downward to raise herself slowly at constant speed? (b) If she increases this force by 10 percent, what will her acceleration be? The mass of the person plus the bucket is 65 kg.</p></li>
</ol>
<p>I'm not 100% sure about this, but here goes nothin':</p>
<ol>
<li>(a) 88 cos 45 = 62.22 N</li>
</ol>
<p>(b) 88 sin 45 = 62.22 N</p>
<p>(c) V2 = V1 + at
(1.5) = 0 + a(2.5)
a = 0.6 ms^-2</p>
<p>F = ma
F = (14.5)(0.6)
F = 8.7 N</p>
<p>f cos 45 = 8.7
f = 8.7/(cos 45)
f = 12.3 N</p>
<p>By the way, you have the Giancoli book don't you? I remember number 32 vividly.. lol..</p>
<p>Yes I do. I have the 5th revised edition. Do you know how to do 30 or 32?</p>
<p>I don't know if this is right, but here's a shot:</p>
<ol>
<li>(a) F = ma
F = 2ma (multiply by two because there are two ropes? I vaguely remember my physics teacher doing this on this problem..)</li>
</ol>
<p>Fg = mg</p>
<p>F = 2(mg)
F = 2(65)(9.8) = 1274 N Not too sure about this...</p>
<p>(b) 1274(1.1) = ma
1401.4 = 65 a
a = 21.56 ms^-2</p>
<p>Okay.. this doesn't seem right lol...</p>
<p>LOL I think that acceleration is a bit too large.</p>
<p>Is there a figure for number 30? I'm not too sure about it, but I'll give it a shot..</p>
<ol>
<li>(a) Ft = (3.0)(9.8) = 29.4 N</li>
</ol>
<p>(b) Ft = m (9.8 + 1.60)
Ft = 3.0 (11.4) = 34.2 N</p>
<p>I don't know how the two buckets are connected to each other. is it by a pully or something? I'm sorry if these are all wrong.. lol.. do any of these make sense to you though?</p>
<p>Yes theres a picture. I see 2 buckets one of top of the other. A string is tying the first bucket to the ceiling. Another string is tying the 2nd bucket to the bottom of the first bucket. Everything is hanging at rest.</p>
<p>You only calculated the tension for 1 chord. How do you calculate the tension for the other chord? Do you simply add the 2 weights for the mass?</p>
<p>Ahh okay.. scratch post 14, I'll re-try this one then!</p>
<p>Yeah, I calculated the tension of the string between the two buckets, to find the tension in the string attached only to the top bucket, add the two masses. To find the tension when it's accelerated upward at 1.6 ms^-2, just do..</p>
<p>Ft = m (9.8 + 1.6)
Ft = 6 (11.4) = 68.4 N</p>
<p>I think this is it..</p>
<p>SORRY...........got 31 wrong.....my mistake</p>
<p>31a) All the Forces in y = ma (a in y direction)
F-6500(9.8) - 12480 = 6500(.6)
F= 80, 080 N</p>
<p>31b) all the Forces in y = ma (a in y direction)
T - Fg = ma
T - 1200(9.8) = 1200(.6)
T=12480 N</p>
<p>thats better....i forgot i got it wrong the first time</p>