Question #9881b
1 Answer
Explanation:
Ideally, you need to have units for all of these quantities, but I'll make the common assumption that we're dealing with
We're asked to find the work done by the tension force on the tram.
To do this, we can use the equation
#ul(W = Fscosphi#
where
-
#W# is the work done by the tension force (what we're trying to find) -
#F# is the magnitude of the tension force (#500# #"N"# ) -
#s# is the displacement (we'll be finding this) -
#phi# is the angle with respect to the horizontal (given as#60^"o"# )
To find the displacement given the acceleration and time, we can use the equation
#ul(s = v_(0x)t + 1/2a_xt^2#
where
-
#s# is the change in position (what we're trying to find) -
#v_(0x)# is the initial velocity (zero assuming it started from rest) -
#t# is the time (#30# #"s"# ) -
#a_x# is the acceleration (#5# #"m/s"^2# )
Plugging in known values:
#s = 0 + 1/2(5color(white)(l)"m/s"^2)(30color(white)(l)"s")^2 = color(red)(ul(2250color(white)(l)"m"#
Therefore, we have
#color(blue)(W) = Fscosphi = (500color(white)(l)"N")(color(red)(2250color(white)(l)"m"))cos(60^"o") = color(blue)(ulbar(|stackrel(" ")(" "5.63xx10^5color(white)(l)"J"" ")|)#
