Question

Question 4 pls my teacher not really teaching my anything in class

Answer 1

C. 4KE
`_(2s)` = KE
`_(6s)` .

Step-by-step explanation:

If we're being asked to compare the kinetic energy of the object at 2s and at 6s, it would be helpful to first evaluate what the amount of kinetic energy at each of these points is.

KE
`_(2s)` - kinetic energy at 2s

The formula for kinetic energy is KE=0.5mv², where m is mass and v is velocity. Our objects mass is 12kg, and from the graph we can see that when the time is 2s, the velocity is 2m/s, so let's sub in 12 for m and 2 for v:

KE
`_(2s)` = 0.5mv²

KE
`_(2s)` = 0.5(12)(2)²

KE
`_(2s)` = 6(4) = 24

So KE
`_(2s)` = 24J

KE
`_(6s)` - kinetic energy at 6s

Again, the formula for kinetic energy is KE=0.5mv², where m is mass and v is velocity. Our objects mass is again 12kg, and from the graph we can see that when the time is 6s, the velocity is 4m/s, so let's sub in 12 for m and 4 for v:

KE
`_(6s)` = 0.5mv²

KE
`_(6s)` = 0.5(12)(4)²

KE
`_(6s)` = 6(16) =96

So KE
`_(6s)` = 96J

Now we know the value of KE
`_(2s)` and KE
`_(6s)`, we can compare them. Let's say that when compared, KE
`_(6s)` is 'x' times bigger than KE
`_(2s)`. Written mathematically, this is:

(x) KE
`_(2s)` = KE
`_(6s)`

Let's sub in our values for KE
`_(2s)` and KE
`_(6s)`:

(x) (24) = 96

x= 96/24 = 4

So since x=4, we can say 4KE
`_(2s)` = KE
`_(6s)` . So our answer is C

Hope this helps. Let me know if you want me to explain anything further :)