Question

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the work done over a distance of 10 cm by (a) the horizontal force, (b) the frictional force, and (c) the net force. (d) What is the change in kinetic energy of the box?

Answer 1

Answer:

(A) Work done by horizontal force will be equal to 3.45 J

(B) Work done by frictional force will be equal to 2.45 J

(C)  Work done by frictional force will be equal to 1 J

(D) Change in kinetic energy will be equal to 1 J

Explanation:

We have given mass of the box m = 5 kg

Acceleration of the box $a=2m/sec^2$

Coefficient of friction $\mu _k=0.5$

Distance s = 10 cm = 0.1 m

(a) Horizontal force will be equal; to

$F=ma+\mu _kmg=5\times 2+0.5\times 5\times 9.8=34.5N$

Now work done $W=Fs=34.5\times 0.1=3.45J$

(b) Frictional force $f=\mu _kmg=0.5\times 5\times 9.8=24.5N$

So work done $W=24.5\times 0.1=2.45j$

(C) Net force will be equal to

Horizontal force - frictional force

So net force = 34.5 -24.5 = 10 N

So work done = 10×0.1 = 1 J

Work done by net force will be equal to change in kinetic energy

So change in kinetic energy will be equal to 1 J

Answer 2

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J