Question: Can you please show all the steps thaks A 40.0-kg box initially at res

Question

Question: Can you please show all the steps thaks

A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find (a) the work done by the applied force, (b) the increase in internal energy in the box-floor system due to friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box.

Explanation / Answer

Frictional force = umg = 0.3*40*9.8 = 117.6 N

Work done by a force = force*displacement * cos(angle between force and displacement)

(a) W by applied force = 130*5*cos(0) = 650 J

(b)Work done by friction converts into internal energy,

W by friction = 117.6 * 5* cos 0= 588 J

(c) Normal force is perpendicular to displacement, so W = 0, since cos90 = 0

(d) Gravitation is also perp to disp. so, W = 0

(e) Change in KE = work done by force - internal energy = 650-588 = 62 J

(f) change in KE = 62 J

1/2 mv^2 = 62 J

v = 1.76 m/s

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