Work and kinetic energy?

Question

A 27.1 kg box initially at rest is pushed 2.56 m along a rough, horizontal floor with a constant applied horizontal force of 129.712 N.
The acceleration of gravity is 9.8 m/s2.

(A) If the coefficient of friction between box and floor is 0.418, find the work done by the applied force. Answer in units of J.

(B) Find the work done by the friction in units of J.

(C) Find the change in kinetic energy of the box in units of J.

(D) Find the the final speed of the box in units of m/s

please and thank you to anyone helps, it's truly appreciated :)

Answer 1

The friction force is umg = (.418)(27.1 kg)(9.8 m/s^2).

(a) This is a little misleading as the work done by the applied force isn't dependent on friction. That work is simply the product of the applied force and the distance moved, or ( 129.712 N ) ( 2.56 m )

(b) Again, work is force multiplied by distance. This time, it's the friction force calculated at the beginning times 2.56 m.

(c) The work done by the applied force you found in (a) is used to overcome friction (b) and to give the box kinetic energy. So the change in kinetic energy is your answer to (a) less your answer to (b).

(d) Once you have the kinetic energy, set it equal to 1/2 m v^2 to find the velocity of the box.

Answer 2

certainly thermal power is kinetic temperature is the diploma of the essential kinetic power of the debris in a pattern yet there is likewise skill power this is power linked with place or some condition an merchandise sitting on a table has skill power with admire to the floor yet there is not any action if the object falls, then that skill power is converted to kinetic chemical bonds have skill power this is on occasion converted to kinetic (in general interior the form of thermal power) as while a gasoline inclusive of methane is burned the debris interior the nucleus of an atom have skill power that is converted to kinetic (many times thermal) power while the debris are rearranged as in fission or fusion of nuclei

Answer 3

N-mg=ma=zero
f(friction force)-F=ma
111.01244-129.712=27.1a
a=-0.690020m/s^2
f=ma=27.1*-0.690020
w=f.s =18.699*2.56=47.870 J

B- f=mu(coeff. of friction)*N
=111.01244
W=f.s
W=284.1918 J
C- k.E(change)=W=1/2m.v.v(intial)-1/2m.v.v(final)
=1/2m(0-v^2(final))47.870
=v(final)=1.8795m/s^2
d- change in K.E IS the work done =47.870J