If a rocket with a mass of #900# #"tons"# vertically accelerates at a rate of #3# #"m/s"^2#, how much power will the rocket have to exert to maintain its acceleration at 2 seconds?
2 Answers
(if given mass is in metric tons)
Explanation:
I'll assume it is
We're asked to find the power output the rocket must have so that it continues this acceleration for
Let's first make the conversion from (metric) tons to newtons:
#900cancel("tonnes")((9807color(white)(l)"N")/(1cancel("tonne"))) = color(red)(ul(8.826xx10^6color(white)(l)"N"#
Its mass is thus
#m = w/g = color(red)(8.826xx10^6color(white)(l)"N")/(9.81color(white)(l)"m/s"^2) = ul(9.00xx10^5color(white)(l)"kg"#
The net force on the rocket gives it an acceleration of
#sumF_y = (9.00xx10^5color(white)(l)"kg")(3color(white)(l)"m/s"^2) = color(orange)(ul(2.70xx10^6color(white)(l)"N"#
So the thrust force exerted by the rocket is
#sumF_y = F_"thrust" - w#
#F_"thrust" = overbrace(color(orange)(2.70xx10^6color(white)(l)"N"))^"net force" + overbrace(color(red)(8.826xx10^6color(white)(l)"N"))^"weight" = color(green)(ul(1.1526xx10^7color(white)(l)"N"#
We can now find the work done by the thrust force using the equation
#ul(W_"thrust" = F_"thrust"·s)color(white)(aa)# (one dimensional)
We can find the displacement
#ul(s = v_0t + 1/2at^2#
where
-
#v_0# is the initial velocity, which is#0# assuming it started from a state of rest -
#t# is the time (#2# #"s"# ) -
#a# is the constant acceleration (#3# #"m/s"^2# )
Plugging in known values, we have
#s = 0 +1/2(3color(white)(l)"m/s"^2)(2color(white)(l)"s")^2#
#s = ul(6color(white)(l)"m"#
The work done by the thrust force is thus
#W_"thrust" = (color(green)(1.1526xx10^7color(white)(l)"N"))(6color(white)(l)"m") = color(purple)(ul(6.92xx10^7color(white)(l)"J"#
Now, we can find the power exerted by the thrust force using the equation
#P_"thrust" = (W_"thrust")/t#
where
-
#W_"thrust"# is the work done by the thrust force (#color(purple)(6.92xx10^7color(white)(l)"J"))# -
#t# is the time maintaining that work (#2# #"s"# )
Thus, we have
#P_"thrust" = (color(purple)(6.92xx10^7color(white)(l)"J"))/(2color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "3.46xx10^7color(white)(l)"W"" ")|)#
The power the rocket must exert is thus
You can redo these conversions if you meant a different type of ton, and the results are
-
short tons:
#color(blue)(3.14xx10^7color(white)(l)"W"# -
long tons:
#color(blue)(3.51xx10^7color(white)(l)"W"#
Explanation:
It is assumed that mass of rocket does not change in the stated 2 seconds. Also air friction is ignored.
Rocket is required to generate power to
- Overcome force due to gravity
- maintain acceleration at
#3ms^-2# for#2s# .
Total upwards acceleration required
Using Newton's Second Law of Motion
Force exerted by rocket
Distance moved by rocket in
#s=ut+1/2at^2#
Inserting given quantities we get
Work done by the force
Power of Rocket