# Question 9881b

Aug 24, 2017

$W = 5.63 \times {10}^{5}$ $\text{J}$

#### Explanation:

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

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 ($500$ $\text{N}$)

• $s$ is the displacement (we'll be finding this)

• $\phi$ is the angle with respect to the horizontal (given as ${60}^{\text{o}}$)

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)

• ${v}_{0 x}$ is the initial velocity (zero assuming it started from rest)

• $t$ is the time ($30$ $\text{s}$)

• ${a}_{x}$ is the acceleration ($5$ ${\text{m/s}}^{2}$)

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"" ")|)#