- #1
Fanman22
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A 391 kg piano slides 3.7 m down a(n) 27° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 7-23). The effective coefficient of kinetic friction is 0.40.
Figure 7-23
(a) Calculate the force exerted by the man.
374N
(b) Calculate the work done by the man on the piano.
-1384J
(c) Calculate the work done by the friction force.
-5054J
(d) Calculate the work done by the force of gravity.
6438J
(e) Calculate the net work done on the piano.
_____J
I got the previous questions correct with the exception of letter "e".
This is what I did...
WORKnet = Wgravity-Wfriction+Wnormal force in x direction-Work man in x direction
My numbers were 6438-5054+6437-1384 = 6437J
The problem must be in the way I'm calculating the normal force in the x direction. I used Fn=Fg*sin27*3.7m=6437J
Thanks in advance for the help.
Figure 7-23
(a) Calculate the force exerted by the man.
374N
(b) Calculate the work done by the man on the piano.
-1384J
(c) Calculate the work done by the friction force.
-5054J
(d) Calculate the work done by the force of gravity.
6438J
(e) Calculate the net work done on the piano.
_____J
I got the previous questions correct with the exception of letter "e".
This is what I did...
WORKnet = Wgravity-Wfriction+Wnormal force in x direction-Work man in x direction
My numbers were 6438-5054+6437-1384 = 6437J
The problem must be in the way I'm calculating the normal force in the x direction. I used Fn=Fg*sin27*3.7m=6437J
Thanks in advance for the help.