Final speed of 12-pack pushed 1.60 m with friction

[rrastam]

HERE IS ANOTHER qUESTION...I GOT THE FIRST PART RIGHT BUT I CAN'T GET PART B RIGHT. PLEASE HELP
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.60 m by a trained dog who exerts a horizontal force with magnitude 37.5 N.
(a) Use the work-energy theorem to find the final speed of the 12-pack if there is no friction between the 12-pack and the floor.
5.28 m/s
(b) Use the work-energy theorem to find the final speed of the 12-pack if the coefficient of kinetic friction between the 12-pack and the floor is 0.30.
? m/s

 

[Pyrrhus]

Work energy Theorem

$$\sum W = \Delta K$$

$$W_{dog} - W_{friction} = \frac{1}{2}mv^2$$

 

[wais]

To solve for part B, we need to incorporate the frictional force into our calculation using the work-energy theorem. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done is equal to the force of the dog (37.5 N) multiplied by the distance the 12-pack is pushed (1.60 m). This work done will result in an increase in the kinetic energy of the 12-pack.

However, we also need to take into account the work done by the frictional force. The work done by friction is equal to the force of friction (μmg) multiplied by the displacement (1.60 m). Therefore, the work done by friction will result in a decrease in the kinetic energy of the 12-pack.

Using the work-energy theorem, we can set up the following equation:

Work done by dog = Work done by friction + Change in kinetic energy

(37.5 N)(1.60 m) = (μmg)(1.60 m) + (1/2)(m)(vf^2)

Substituting in the given values, we get:

(37.5 N)(1.60 m) = (0.30)(4.30 kg)(9.8 m/s^2)(1.60 m) + (1/2)(4.30 kg)(vf^2)

Solving for vf, we get:

vf = √[(37.5 N)(1.60 m) - (0.30)(4.30 kg)(9.8 m/s^2)(1.60 m)] / (1/2)(4.30 kg)

vf = 2.98 m/s

Therefore, the final speed of the 12-pack with a coefficient of kinetic friction of 0.30 is 2.98 m/s.