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# Answer to Question #5010 in Mechanics | Relativity for Melanie

Question #5010
A 2.10 102-N force is pulling an 70.0-kg refrigerator across a horizontal surface. The force acts at an angle of 17.0&deg; above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 8.00 m. Find the work done by the kinetic frictional force.
1
2011-11-10T10:26:28-0500
Let&#039;s make some denotations:
F = 102 N
M = 70.0 kg
&alpha; = 17&deg;
&mu; = 0.200
d = 8 m

The horizontal projection of the pulling force is
Fh = F*cos&alpha; = 102*cos(17) = 97,543 N.
Fv = F*sin&alpha; = 102*sin(17) = 29,822 N.
The friction force is:
Ffr = Fn*&mu; = (M*g-Fv)*&mu; = (102*9.8-29,822)*0.2 = 193,956 N.
Here Fn is the normal reaction force.
So, we see that for given conditions the pulling force is too weak to displace the refrigerator. Please check the input values, maybe the friction coefficient is less or refrigirator is lighter?

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