Finding acceleration during a change in momentum

[FaraDazed]

Homework Statement

A block of mass of 60kg is at rest on a smooth horizontal surface and is then pushed with a constant force of 126N for 10 seconds.

Ignoring friction and air resistance;

A. calculate the acceleration of the block during the 10 seconds
B. calculate the velocity after 10 seconds.

Homework Equations

F=ma
Impulse=Force x time
Ns=mv-mu
v=u+at?

The Attempt at a Solution

Part A:

My first attempt A:
126x10=1260
1260/60 = 21ms^2

My second attempt after think i made it more complicated is:
F=ma
126=60xa
a=126/60=2.1ms^2Part B:

First attempt
Ns=mv-mu
1260=(60xv)-(60-0)
1260+60=60xv
1320=60xv
v=1320/60= 22ms

Second Attempt:
126x10=1260
1260/30=10.5

Further Discussion:

I am assuming from the question that when they say "calculate the velocity after 10 seconds" they mean the final velocity (v)? If so then couldn't I use the motion equation v=u+at to find Part B if I know that Part A is right?

 

[SHISHKABOB]

In your first attempt at part A it seems like you multiplied the force with the duration of the push and then divided that by the mass and you say that this is equal to the acceleration.

or

\frac{Ft}{m} = a

if you just look at the units, you should be able to see that the left side and the right side of that equation are not equal to each other, so this cannot be right

in your second attempt you do

\frac{F}{m} = a

which follows from F = ma, which makes sense

basically, if the force is constant, the length of time for which the object has the force applied to it doesn't matter

it doesn't matter if it's pushed for 1 second, 1 hour or 1 year, the acceleration will always be the same, because the force is equal to the mass times the acceleration, and there is no dependency on time.
You are correct that in part B they are asking for the final velocity of the object. If the object has been accelerating at a constant value for ten seconds, then it will have a final velocity given by the equation v = u + at. There's no need to worry about the mass of the object or the force acting on it; you only need to worry about the value of the acceleration.

 

[FaraDazed]

Thank you, you helped a lot. :)