Question #9881b

1 Answer
Aug 24, 2017

Answer:

#W = 5.63xx10^5# #"J"#

Explanation:

Ideally, you need to have units for all of these quantities, but I'll make the common assumption that we're dealing with #60^"o"#, #500# #"N"#, and #5# #"m/s"^2#.

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"" ")|)#