A 36.0 kg box initially at rest is pushed 5.75 m along a rough, horizontal floor

Question

A 36.0 kg box initially at rest is pushed 5.75 m along a rough, horizontal floor with a constant applied horizontal force of 135 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force J (b) the increase in internal energy in the box-floor system due to friction J (c) the work done by the normal force J (d) the work done by the gravitational force J (e) the chance in kinetic energy of the box J (f) the final speed of the box m/s

Explanation / Answer

a) Work done by the applied force W1 = F*S*cos(theta)

theta is the angle between Force F and displacement S

then W1 = F1*S*cos(0) = 135*5.75*cos(0) = 776.25 J

b) work done by the friction = change in internal energy on the box floor system due to friction = F2*S = mu_k*m*g*S*cos(0) =0.3*36*9.8*5.75*cos(0) = 608.58 J

C) Work done by the Normal force W3 = F3*S*cos(90)

here theta = 90 degrees since the normal force is perpendicular to the surface i.e distance S

then W3 = 0 J

Work done by the gravitational force W4 = F4*S*cos(90) = 0 J ,since the gravitational force is perpendiular to the displacement

e) According to work energy theorem ,work done by the net force = change in kinetic energy

Work done by the applied force + frictional force = change in kinetic energy

776.25-608.58 = change in kinetic energy

change in kinetic energy = 167.67 J

f) change in kinetic energy = Final KE - initial KE

since the box initially at rest then its initial kinetic energy is zero

then

167.67 = 0.5*m*V^2

167.67 = 0.5*36*v^2

v = 3.05 m/sec

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