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Title: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shell is Post by: Dannyy on Sep 1, 2015 A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shell is fired from the cannon. The cannon recoils with a speed of 12 m/s. How far will the shell travel in 30 seconds? Please show work as I have an answer but believe I may be using the wrong equation.
Consider a tank that weighs 980 N at rest on a conveyor belt. The belt on the conveyor is moving horizontally at 1 ft/sec. The conveyor belt is driven by a motor that provides a time dependent force, in Newtons, described as F(t) = 1000 + 0.25t. (A) How much work was done on the tank? (B) What was the change in potential energy of the tank? Again, please show work. Title: Re: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shel Post by: bio_man on Sep 1, 2015 A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shell is fired from the cannon. The cannon recoils with a speed of 12 m/s. How far will the shell travel in 30 seconds? Please show work as I have an answer but believe I may be using the wrong equation. First attachment. Consider a tank that weighs 980 N at rest on a conveyor belt. The belt on the conveyor is moving horizontally at 1 ft/sec. The conveyor belt is driven by a motor that provides a time dependent force, in Newtons, described as F(t) = 1000 + 0.25t. (A) How much work was done on the tank? (B) What was the change in potential energy of the tank? Again, please show work. Second attachment. Title: Re: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shel Post by: Dannyy on Sep 1, 2015 Bioman,
I appreciate the help but I also found these answers on a different website. When I do these problems I come up with different answers. From what you know, are these answers correct? Please let me know what you think. Title: Re: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shel Post by: bio_man on Sep 1, 2015 I appreciate the help but I also found these answers on a different website. Which website, if you don't mind me asking? In the second question, we are taking the integral of the original equation, then substituting the velocity. Are you taking calculus? The first question is straight forward, plug and play. What answers did you find? Title: Re: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shel Post by: location on Sep 1, 2015 They were on chegg
Title: Re: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shel Post by: Dannyy on Sep 1, 2015 Bioman,
This is what I got for the first one. velocity of the conveyor belt=1 ft/s=0.3048 m/s power=force*velocity power in terms of time =(1000+0.25*t)*0.3048=304.8+0.0762*t then work done=integral of power*dt =integral of (304.8*dt+0.0762*t*dt) =304.8*t+0.0381*t^2 J b) as the belt is moving horizontally, there is no change in elevation. hence change in potential energy=0 Used Conservation of Momentum (1110+25)*15 = (-1110*12)+(25*v) v = 1213.8 m/s distance travelled is v*t = 1213.8*30 = 36414 m/s Title: Re: A 1,110 kg cannon loaded with a 25.0 kg shell is rolling down a 45 degree hill at 15 m/s. A shel Post by: dtimmons95 on Sep 1, 2015 (1110+25)*30 = (25*v1) - (1110*12)
v1 = 1213.8 m/s distance = v1*t= 36414 m |