Question #37343

A 370 N force is pulling an 85.1-kg refrigerator across a horizontal surface. The force acts at an angle of 22.0 ° above the surface. The coefficient of kinetic friction is 0.237, and the refrigerator moves a distance of 9.44 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

Expert's answer

A 370 N force is pulling an 85.1-kg refrigerator across a horizontal surface. The force acts at an angle of 22.0° above the surface. The coefficient of kinetic friction is 0.237, and the refrigerator moves a distance of 9.44 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

Solution

Here we have


m=85.1kgF=370Nα=22.0μ=0.237S=9.44m\begin{array}{l} m = 85.1\,kg \\ F = 370\,N \\ \alpha = 22.0{}^{\circ} \\ \mu = 0.237 \\ S = 9.44\,m \\ \end{array}


We choose the x-axis parallel to horizontal surface, y-axis orthogonal to horizontal surface.

In y-axis we have second Newton's equation law P+Fsinαmg=0P + F \sin \alpha - mg = 0, where PP is reaction of surface. From hence P=mgFsinα=695.4NP = mg - F \sin \alpha = 695.4\,N,

We have friction force Ff=μPF_f = \mu P.

The pulling force done work A=FS=FScosα=3239fA = FS = FS \cos \alpha = 3239\,f (here α\alpha is angel between pulling force and translation).

The friction forces done work AF=FfS=μPS=1556fA_F = F_f S = \mu P S = 1556\,f (because friction force is antiparallel to translation).

Answer:

a) The pulling force done work A=3239fA = 3239\,f.

b) The friction forces done work AF=1556fA_F = 1556\,f

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