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A 10 kg object experiences a horizontal force which causes it to accelerate at
5 m/s^{2}, moving it a distance of 20 m, horizontally. How much work is
done by the force?

The magnitude of the force is given by F = ma = (10)(5) = 50 N. It acts over a
distance of 20 m, in the same direction as the displacement of the object,
implying that the total work done by the force is given by W = Fx = (50)(20) = 1000
Joules.

Problem :

A ball is connected to a rope and swung around in uniform circular motion. The
tension in the rope is measured at 10 N and the radius of the circle is 1 m. How
much work is done in one revolution around the circle?

Recall from our study of uniform circular motion that
centripetal force is always directed
radially, or toward the center of the circle. Also, of course, the displacement
at any given time is always tangential, or directed tangent to the circle:
Clearly the force and the displacement will be perpendicular at all times. Thus
the cosine of the angle between them is 0. Since W = Fx cosθ, no work is
done on the ball.

Problem :

A crate is moved across a frictionless floor by a rope THAT is inclined 30
degrees above horizontal. The tension in the rope is 50 N. How much work is done
in moving the crate 10 meters?

In this problem a force is exerted which is not parallel to the displacement of
the crate. Thus we use the equation W = Fx cosθ. Thus

W = Fx cosθ = (50)(10)(cos 30) = 433 J

Problem :

A 10 kg weight is suspended in the air by a strong cable. How much work is done,
per unit time, in suspending the weight?

The crate, and thus the point of application of the force, does not move. Thus,
though a force is applied, no work is done on the system.

Problem :

A 5 kg block is moved up a 30 degree incline by a force of 50 N, parallel to the
incline. The coefficient of kinetic friction between the block and the incline
is .25. How much work is done by the 50 N force in moving the block a distance
of 10 meters? What is the total work done on the block over the same distance?

Finding the work done by the 50 N force is quite simple. Since it is applied
parallel to the incline, the work done is simply W = Fx = (50)(10) = 500 J.

Finding the total work done on the block is more complex. The first step is to
find the net force acting upon the block. To do so we draw a free body diagram:

Because of its weight, mg, the block experiences a force down the incline of
magnitude mg sin 30 = (5)(9.8)(.5) = 24.5 N. In addition, a frictional
force is felt opposing the motion, and thus down the incline. Its magnitude is
given by F_{k} = μF_{N} = (.25)(mg cos 30) = 10.6 N. In addition, the normal
force and the component of the gravitational force that is perpendicular to the
incline cancel exactly. Thus the net force acting on the block is: 50 N -24.5 N -10.6 N = 14.9 N, directed up the
incline. It is this net force that exerts a ìnet workî on the block. Thus the
work done on the block is W = Fx = (14.9)(10) = 149 J.