Work and kinetic friction problem

[kmotoao]

i have two questions that needs help on. thanks in advance

1. You pull a box across the floor with a force of 425 N. The coefficient of kinetic friction is 0.305. The mass of the crate is 125 kg. Angle q = 35.0°. Find: (a) the acceleration of the box and (b) the amount of work done in moving the crate a distance of 3.50 m.

2. A 2.5 kg box slides across the flat surface of a table. The coefficient of kinetic friction for the table/box is 0.295. The box is attached to a light string that passes over a low friction pulley and is connected to a 3.0 kg mass that is hanging vertically. (a) find the acceleration of the system (b) find the velocity of the 2.5 kg box after it has been dragged 0.25 m if its initial velocity was 0.25 m/s, and (c) find the kinetic energy of the box at this point.

 

[ek]

It's late, and I'm going to bed, but I did the first one. The most important thing is to realize that the Fn in Ff = uFn is not just mg but mg-Fsinx, because there is a component of that force pulling in the y direction, effectively lessening the normal force that friction acts on. So:

Ff = uFn

where Fn = mg - Fsinx

Ff = u(mg-Fsinx)

and

F' = Fcosx - Ff

then

F' = Fcosx - u(mg-Fsinx)

and

a = Fcosx - u(mg-Fsinx)/m

And for the work it's just Fcosx*d.

 

[wais]

For question 1:

(a) To find the acceleration of the box, we can use the formula a = Fnet/m, where Fnet is the net force acting on the box and m is the mass of the box. We know that the net force is equal to the applied force minus the force of kinetic friction, so Fnet = 425 N - (0.305)(125 kg)(9.8 m/s^2) = 379.875 N. Plugging this into the formula, we get a = 379.875 N/125 kg = 3.039 m/s^2. Therefore, the acceleration of the box is 3.039 m/s^2.

(b) To find the work done in moving the crate a distance of 3.50 m, we can use the formula W = Fd, where W is work, F is the applied force, and d is the distance moved. Plugging in the values, we get W = (425 N)(3.50 m) = 1487.5 J. Therefore, the amount of work done in moving the crate a distance of 3.50 m is 1487.5 J.

For question 2:

(a) To find the acceleration of the system, we can use the formula a = (m1g - m2g - Fk)/m1, where m1 is the mass of the box, m2 is the hanging mass, g is the acceleration due to gravity, and Fk is the force of kinetic friction. Plugging in the values, we get a = [(2.5 kg)(9.8 m/s^2) - (3.0 kg)(9.8 m/s^2) - (0.295)(2.5 kg)(9.8 m/s^2)]/2.5 kg = -0.784 m/s^2. Therefore, the acceleration of the system is -0.784 m/s^2.

(b) To find the velocity of the 2.5 kg box after being dragged 0.25 m, we can use the formula vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance moved. Plugging in the values, we get vf^2 = (0.25 m/s)^2 + 2(-0.784 m