Physics Work and Energy Question!

Question

A boy pulls a 10.0 kg book rack at rest on a smooth floor with a constant horizontal force of 1.2N starting from rest.

Compute the change in kinetic energy of the box after the boy has pulled it a distance of 5.0m

AND

compute the final velocity of the box after the boy has pulled it a distance of 5.0m.

 

I have no idea where to even start. Help would be massively appreciated. Thank you in advance. 

Answer 1

Hello Haizai,

 

To start with, let's calculate the acceleration of the box, knowing that

F=ma

m=10.0 kg

f=1.2N

 

We can solve for a

a=F/m

 =1.2kg/10.0n

 = .12m/s²

 

Now that we know acceleration, we can solve for the final velocity using:

v_f²=v₀²+2ax

v_f (final velocity) =unknown

v₀ (Initial velocity) = 0

a = .12m/s²

x (change in position) = 5.0m

 

v_f² = 0+2(.12)(5.0)

v_f² = 1.2

v_f = √1.2

v_f = 1.1m/s

 

Finally, kinetic energy is:

K=1/2mv²

K=1/2(10)(1.1)²

K=6J

Answer 2

Haizai,

 

Since you are studying work & energy, you might want to consider using the work-energy theorem for your solution.

 

Since the question states that the floor is smooth, we don't have to consider losses due to friction.

 

The Work-Energy Theorem says that the work done on an object is equal to the change in it kinetic and/or potential energy.  Since the box is staying on the smooth floor, no work is being done against gravity and we don't have to consider any change in potential energy.  This means all the work goes into changing the box's kinetic energy.  Hence:

 

Work = F·d = ½ m·v² = kinetic energy

 

Rearranging gives:

 

v² = (2 F·d)/m ⇒ v = [(2 F·d)/m)]½

 

You get the same answers that Anthony showed you, but with a bit faster and with less computation.