(a) 69.3 J
The work done by the applied force is given by:
[tex]W=Fd cos \theta[/tex]
where:
F = 25.0 N is the magnitude of the applied force
d = 3.20 m is the displacement of the sled
[tex]\theta=30^{\circ}[/tex] is the angle between the direction of the force and the displacement of the sled
Substituting numbers into the formula, we find
[tex]W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J[/tex]
(b) 0
The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.
(c) 69.3 J
According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:
[tex]\Delta K = W[/tex]
where
[tex]\Delta K[/tex] is the change in kinetic energy
W is the work done
Since we already calculated W in part (a):
W = 69.3 J
We therefore know that the change in kinetic energy of the sled is equal to this value:
[tex]\Delta K=69.3 J[/tex]
(d) 4.9 m/s
The change in kinetic energy of the sled can be rewritten as:
[tex]\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex] (1)
where
Kf is the final kinetic energy
Ki is the initial kinetic energy
m = 5.50 kg is the mass of the sled
u = 0 is the initial speed of the sled
v = ? is the final speed of the sled
We can calculate the variation of kinetic energy of the sled, [tex]\Delta K[/tex], after it has travelled for d=3 m. Using the work-energy theorem again, we find
[tex]\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J[/tex]
And substituting into (1) and re-arrangin the equation, we find
[tex]v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s[/tex]
