Problem #1
What is the work done by friction in slowing a 10.5kg block traveling at 5.85 m/s to a complete stop in a distance of 9.65 m? What is the kinetic coefficient of friction?Answer:
Since we are asked for the work done and have a change in speed, we make use of the generalized WorkEnergy Theorem. Since the height of the block does not change, there is only a change in kinetic energy.
Wfric = ΔE = Kf – Ki
= ½m[(vf)2 – (v0)2]
= ½(10.5kg)[(0)2 – (5.85 m/s)2]
Wfric = 179.67 J
To find μ, we need to know the force and the angle it makes with the displacement. To find forces, we draw a FBD and use Newton's Second Law.
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| Fig.1 |
Wfric = fkΔx = μkmgΔx.
Rearranging this yields
μk = Wfric/mgΔx = 0.18
Problem #2
A 50.0N force is applied horizontally to a 12.0kg block which is initially at rest. After traveling 6.45 m, the speed of the block is 5.90 m/s. What is the coefficient of kinetic friction?
Answer:
Since the problem involves a change is speed, we make use of the Generalized Work-Energy Theorem:
WNC = ΔE = Kf – Ki
= ½m[(vf)2 – (v0)2]
WNC = ½m(vf)2 .
There are two nonconservative forces in this problem, friction and the applied force. The work done by friction is given by Wfric = -fkΔx. The work done by the applied force is WF = FΔx
FΔx – fkΔx = ½m(vf)2 (1)
(1) To find out more about fk , we draw a FBD and use Newton's Second Law.
In Fig.1, the second equation gives N = mg and we know fk = μkN, so fk = μkmg. Thus
Wfric = μkmgΔx.
Combining this result with equation (1), we get
(F – μkmg)Δx = ½m(vf)2 .
Rearranging yields an expression for μk
μk = [FΔx – ½m(vf)2]/mgΔx.
Using the given values, we find μk = 0.15 .
Problem #3
A 40 kg box initially at rest is pushed 5.0m along a rough horizontal floor with a constant applied horizontal force of 130N. If the coefficient of friction between the box and floor is 0.30, find (a) the work done by the applied force, (b) the energy lost due to friction, (c) the change in kinetic energy of the box, and (d) the final speed of the box.
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| Fig.2 |
(a) The motion of the box and the forces which do work on it are shown in Fig. 2(a). The (constant) applied force points in the same direction as the displacement. Our formula for the work done by a constant force gives
Wapp = Fdcos θ = (130 N)(5.0 m) cos 00 = 6.5 × 102 J
The applied force does 6.5 × 102 J of work.
(b) Fig. 2(b) shows all the forces acting on the box. The vertical forces acting on the box are gravity (mg, downward) and the floor’s normal force (N, upward). It follows that N = mg and so the magnitude of the friction force is
ffric = μN = μmg = (0.30)(40 kg)(9.80 m/s2) = 1.2 × 102 N
The friction force is directed opposite the direction of motion (θ = 1800) and so the work that it does is
Wfric = Fdcos θ = ffric d cos 1800 = (1.2 × 102 N)(5.0 m)(−1) = −5.9 × 102 J
or we might say that 5.9 × 102 J is lost to friction.
(c) Since the normal force and gravity do no work on the box as it moves, the net work done is
Wnet = Wapp +Wfric = 6.5 × 102 J − 5.9 × 102 J = 62 J
By the work–Kinetic Energy Theorem, this is equal to the change in kinetic energy of the box:
ΔK = Kf − Ki = Wnet = 62 J
(d) Here, the initial kinetic energy Ki was zero because the box was initially at rest. So we have Kf = 62 J. From the definition of kinetic energy, K = ½ mv2, we get the final speed of the box:
v2f = 2Kf/m = 2(62 J)/(40 kg) = 3.1 m2/s2
so that
vf = 1.8 m/s
Problem #4
A crate of mass 10.0kg is pulled up a rough incline with an initial speed of 1.50 m/s . The pulling force is 100N parallel to the incline, which makes an angle of 200 with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00m. (a) How much work is done by gravity? (b) How much energy is lost due to friction? (c) How much work is done by the 100N force? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00m?
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| Fig.3 |
(a) We can calculate the work done by gravity in two ways. First, we can use the definition:
W = F · d. The magnitude of the gravity force is
Fgrav = mg = (10.0 kg(9.80 m/s2) = 98.0N
and the displacement has magnitude 5.00 m. We see from geometry (see Fig. 3(b)) that the angle between the force and displacement vectors is 1100. Then the work done by gravity is
Wgrav = Fdcos θ = (98.0 N)(5.00 m) cos 1100 = −168 J
Another way to work the problem is to plug the right values. From simple geometry we see that the change in height of the crate was
Δy = (5.00 m) sin 200 = +1.71m
Then the work done by gravity was
Wgrav = −mgΔy = −(10.0 kg)(9.80 m/s2)(1.71 m) = −168 J
(b) To find the work done by friction, we need to know the force of friction. The forces on the block are shown in Fig. 3(a). As we have seen before, the normal force between the slope and the block is mg cos θ (with θ = 200) so as to cancel the normal component of the force of gravity. Then the force of kinetic friction on the block points down the slope (opposite the motion) and has magnitude
fk = μkN = μmg cos θ = (0.400)(10.0 kg)(9.80 m/s2) cos 200 = 36.8 N
This force points exactly opposite the direction of the displacement d, so the work done by friction is
Wfric = fkd cos 1800 = (36.8 N)(5.00 m)(−1) = −184 J
(c) The 100N applied force pulls in the direction up the slope, which is along the direction of the displacement d. So the work that is does is
Wappl = Fdcos 00 = (100 N)(5.00 m)(1) = 500. J
(d) We have now found the work done by each of the forces acting on the crate as it moved: Gravity, friction and the applied force. (We should note the the normal force of the surface also acted on the crate, but being perpendicular to the motion, it did no work.) The net work done was:
Wnet = Wgrav + Wfric + Wappl = −168 J − 184 J + 500 J = 148 J
From the work–energy theorem, this is equal to the change in kinetic energy of the box:
ΔK = Wnet = 148 J.
(e) The initial kinetic energy of the crate was
Ki = ½ (10.0 kg)(1.50 m/s)2 = 11.2J
If the final speed of the crate is v, then the change in kinetic energy was:
ΔK = Kf −Ki = ½ mv2 − 11.2 J
Using our answer from part (d), we get:
ΔK = ½ mv2 − 11.2 J = 148J
v2 = 2(159 J)/m
So then:
v2 = 2(159 J)/(10.0kg) = 31.8 m2/s2
v = 5.64 m/s
The final speed of the crate is 5.64 m/s
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