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

The change in the kinetic energy is 1J.

Let us note that the net force is obtained from;

Net force = ma = 5 Kg × 2.0 m/s2 = 10 N

Frictional force;

F =  μKmg = 0.50 × 5kg × 9.8 ms-2 = 24.5 N

The applied force =

10 N = Fa - 24.5 N

Fa = 10 N + 24.5 N

Fa = 34.5 N

Work done = Work done by applied force + work done by friction

Work done = (34.5 N × 0.1) - (24.5 N  × 0.1) = 3.45 - 2.45 = 1.0 J

Kinetic energy of the box = 1/2 mv^2

v^2 = u^2 + 2as

Since u = 0

v^2 = 2as

v = √2 × 2 × 0.1

v = 0.63 m/s

ΔKE = 0.5 × 5 Kg × (0.63)^2 = 1 J

Learn more about kinetic energy: https://brainly.com/question/731147?