Question

PLEASE HELP
A 7.5 kg block initially at rest is pulled to the
right along a horizontal, frictionless surface
by a constant, horizontal force of 19.6 N.
Find the speed of the block after it has
moved 2.4 m.
Answer in units of m/s

I got 1.77 but it's wrong?

Answer 1

you first need to work out your acceleration before you can work out your velocity

Answer 2

U(initial velocity) = 0 ; F = 19.6N; S = 2.4m; M = 7.5kg
calculate V = ?

Answer 3

Actually I just used the equation \[k=.5mv^2\] :)

Answer 4

F = MA

19.6N = 7.5kg(a)
to get the (a) on it own do the following:
\[\frac{ 19.6 }{ 7.5 } = \frac{ 7.5 }{ 7.5 }a\]
2.613 = a
so
a = 2.613 m/s

now we can calculate velocity/speed:
\[s = \frac{ v ^{2} - u 2 }{ 2a }\]
\[2.4m = \frac{ v ^{2} - \left( 0 \right)^{2} }{ 2(2.613) }\]
\[2.4m = \frac{ v ^{2} }{ 5.226 }\]
now to have \[v ^{2}\] alone you must * both sides by \[\frac{ 1 }{ 5.226 }\]
when you canceled out 5.226 on the right side , multiply on the left, which will leave you with :
10.452 = \[10.452 = v ^{2}\]
now take the \[\sqrt{10.452} = \sqrt{v ^{2}}\]
which will leave you with the answer...
\[3.233 = v\]

Answer 5

did you get the same answer as mine?

Answer 6

I got 3.54 but pretty close :) thanks

Answer 7

cool. thanks