Question

Question #9881b

Answer 1

`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"" ")|)`