- #1
TwinGemini14
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Suppose that a 50kg block slides along a horizontal surface where the coefficient of kinetic friction between the block and the surface is 0.60. A force F = 400 N is now applied where the angle of the force above horizontal is 20°. (Like a string pulling the block from the right 20 degrees above the horizontal)
What is the net work done on the block after the block is pulled 15 meters?
A) 1.54 * 10^3 J
B) 2.45 * 10^3 J
C) 2.75 * 10^3 J
D) 3.00 * 10^3 J
E) 3.29 * 10^3 J
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The force applied across the horizontal is Fcos(20) = 400*cos(20) = 375.88N.
The frictional force opposing this is (0.6)(50)(-9.8) = -294N.
So the work done by the applied force is W = Fd = (375.88)(15) = 5638.2J.
The work done by the frictional force is W = Fd = (-294)(15) = -4410J.
The net force is then... W = (5638.2)+(-4410) = 1228.2J.
I got 1.23 * 10^3J ... but that is not a choice. Where did I go wrong? Can anybody help me out here?
What is the net work done on the block after the block is pulled 15 meters?
A) 1.54 * 10^3 J
B) 2.45 * 10^3 J
C) 2.75 * 10^3 J
D) 3.00 * 10^3 J
E) 3.29 * 10^3 J
----
The force applied across the horizontal is Fcos(20) = 400*cos(20) = 375.88N.
The frictional force opposing this is (0.6)(50)(-9.8) = -294N.
So the work done by the applied force is W = Fd = (375.88)(15) = 5638.2J.
The work done by the frictional force is W = Fd = (-294)(15) = -4410J.
The net force is then... W = (5638.2)+(-4410) = 1228.2J.
I got 1.23 * 10^3J ... but that is not a choice. Where did I go wrong? Can anybody help me out here?