physics kinetics question pt 2, please help :)

Question

A shopper exerts a 165 Newton force straight upward to lift a bag of groceries that has a mass of 9.5 kg. Starting from rest, the groceries are lifted from the ground to a height of 1.38 meters. Find the resulting KE of the groceries.
In your answer:
give precision to 0.1.

Answer 1

Hi Lauren!

 

You can also get at this one through the work-(kinetic) energy theorem, which states that the net work done on a system equals its change in kinetic energy:

 

W_net = ΔKE = KE_f - KE_i

 

So we just have to calculate the work done by each force in the situation described, add them up, and use that to determine the kinetic energy change.

 

First, since the bag starts at rest, its initial kinetic energy must be zero (because having kinetic energy depends on having a velocity).  Hence, we have:

 

W_net = KE_f - 0 = KE_f

 

So, in this case, the net work equals the final kinetic energy.

 

To calculate the work done by each force, we use the definition:

 

W = Fdcosα

 

where

 

F = force doing the work'

d = displacement over which the worked-on object moves

α = angle between the direction of the force and the direction of the displacement.

 

For this case, we have two forces doing work:

 

1.  the applied lifting force of the shopper

2.  gravity

 

Let's calculate the work done by each one (keeping the precision of results at 0.1):

 

1.  The lifting force is straight upward, as is the displacement.  Thus, the two vectors are parallel, and the angle between them is α = 0º.  cos(0) = 1, so we have:

 

W_L = F_Ldcos(α) = (165 N)(1.38 m)(1) = 227.7 J

 

2.  Gravity pulls directly downward, so it points directly opposite the displacement, with a force F_g = mg.  Thus, the angle between the two vectors is 180º.  cos(180º) = -1, so we have:

 

W_g = F_gdcosα = (mg)(d)(-1) = -(9.5 kg)(9.8 m/s²)(1.38 m) = -128.5 J

 

Thus, 

 

W_net = W_L + W_g = 227.7 J - 128.5 J = 99.2 J = KE_f

 

I hope this helps!  Just let me know if you have any questions about what I did.

Answer 2

Force upward = 165

Weight acting down = 9.5*9.8

Net force upward = 71.90

Acceleration upward = 71.90 / 9.5 = 7.57 m / sec / sec

 

VF² = VI² + 2as

VF^{2  = 0 + 2 * 7.57 * 1.38 = 20.8932}

Final KE = 1/2 mv² = 0.5 * 9.5 * 20.8932 =99.24 joules