q11

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General Note

In general we would write for the Work-Energy theorem:

W_EXT + W_g + W_s + W_fk = ½mv₂² - ½mv₁². Here, W_EXT = work done by external force. W_EXT can include the work done by the normal force or the tension force if those forces are aligned in directions of non-zero work. Often the work done by the normal force is zero because the motion is perpendicular to that force's direction. W_g= gravitational work, W_s= spring work and W_fk= friction work.   In various problems, these terms may or may not be zero.

In general we could also write an equivalent statement from Conservation of Energy:

½mv₁² + mgy₁+ ½kx₁² + W_EXT =   ½mv₂² + + mgy₂+ ½kx₂²   + HEAT,

where mgy = U_{g ,} ½kx²= U_s , W_EXT is the external force work and HEAT = - W_fk .   If the friction force is constant, then the HEAT = +f_k*D since W_fk = - f_k*D in that case. In various problems, these terms may or may not be zero.

1. AB cos Ø is the formula for the dot product, where A and B are vector magnitudes .

30.

(a) Work = FDcos0, where F = mg and D is given.
(b) Power = Fvcos0 = Fv , where F = mg and v = D/t

35.

(a) F_x = ma_x. Find the x-component of acceleration from Ch. 2 theory: D = ½a_xt², since the sprinter begins from rest .
(b) Find Power = Fvcos0 = F_xv_x at various given points in time; you can find the x-component of velocity by differentiating the expression for acceleration.

41.
(a) W_g = Fhcos180 = -FD = -mgh.
(b) ma_y = pos - neg
ma_y = T - mg
Solve for T then compute W_T = Thcos0 = Th.
(c) and (d) Use Work Energy Theorem, which says the net work ("sum of the works" ) equals the change in kinetic energy. The are only two forces that do work, gravity and tension:
W_NET = W_g + W_T = ½mv₂² - ½mv₁² . The first speed is zero. Find the second kinetic energy and speed.

42.
(a) Net Work = change in kinetic energy = ½mv₂² - ½mv₁²
(b) F = ma, where a = acceleration magnitude. Get a from Chapter 2 theory (kinematics) . Use final speed and the distance for that. The motion begins from rest. Or simply divide answer to (a) by D.
(c) Fvcos0 = Fv , where v is the given maximum speed magnitude, i.e. final speed.

43.

Net Work = change in kinetic energy.
W_g + W_EXT = ½mv₂² - ½mv₁², where the left hand side is the sum of gravitational and external force work. F = magnitude of the external force. Now the motion is up the incline and the force of magnitude F is horizontal. Thus the angle between motion and force is 20 degrees. W_EXT = F*L*cos20. L is the length traveled along the incline which you can compute using the vertical height h and the sin of the angle. Note that the gravitational work is negative. Find the second speed assuming the crate starts from rest.

44.
You can use the Work energy Theorem:
Net Work = change in kinetic energy = ½mv₂² - ½mv₁² .
W_g + W_EXT = ½mv₂² - ½mv₁² , where W_g = work by gravity and W_EXT = work by horizontal force of magnitude F.
Now, the angle between the horizontal force and motion is 180 - 20 = 160 degrees. That's because the force opposes the motion. Thus, W_EXT = FDcos160 = -FDcos20, where D = distance moved along incline. D = 50 /sin20 meters from the geometry of the problem. Wg = +mgh since the skier moves from high to low. Solve for the second speed. Note the first speed is zero.

Note in general we would write: W_EXT + W_g + W_s + W_fk = ½mv₂² - ½mv₁². In this case, the normal force work = 0 since that force is perpendicular to the motion. So the work done by the external force is only due to the force of magnitude F. W_s = 0 because the problem is spring-less. W_fk = 0 because the problem is frictionless.

46.
W_g + W_s +   W_fk = ½mv₂² - ½mv₁². Note that W_EXT = 0; there is no external force. The normal force work = 0 since the force is perpendicular to the motion.

I've labeled all the main forces as subscripts. The gravitational work = 0 since the force is perpendicular to the horizontal motion. W_s =   ½kx₁² - ½kx₂²    where x₂ = 0. The first value of x is - 0.20 m. W_fk = -µ_kND, where D is the distance on the rough part only and N is the normal force.

Note that v₁ = v₂ = 0. Solve for D.

47. W_g + W_s +   W_fk = ½mv₂² - ½mv₁². Note that W_EXT = 0; there is no external force. The normal force work = 0 since the force is perpendicular to the motion.

I've labeled all the main forces as subscripts. The gravitational work = 0, since the force is perpendicular to the horizontal motion. W_s = 0 because the problem is spring-less. W_fk = -µ_kN*D, where D is the distance on the rough surface and N is the normal force.

Note that v₂ = 0. and the first speed is given. Solve for D.

49. You could solve this problem either through the work-energy theorem or conservation of energy. I will provide both inter-related hints:

Conservation of energy :
½mv₁² + mgy₁+ ½kx₁² + W_EXT =   ½mv₂² + + mgy₂+ ½kx₂²   + HEAT,

where mgy = U_{g ,} ½kx²= U_s , W_EXT is the external force work and HEAT = - W_fk .   Since the friction force is constant,   the HEAT = +f_k*D since W_fk = - f_k*D.

(a) W_EXT = 0; the normal force work = 0 since the force is perpendicular to the motion. The first speed is zero. The first y-value is zero. x₁= - 0.30 m. The second x value is zero since the package has lost contact with the spring, which releases the package when un-deformed. You are given the second y-value. No friction ; thus HEAT = 0. Find the second speed value.
(b) The conditions are the same as part (a) except that now friction work and thus heat are not zero because of motion along the ramp's sticky section .
½kx₁² =   ½mv₂² + + mgy₂+ HEAT, where the HEAT = +f_k*D since W_fk = - f_k*D. Note: D is given. Recall
f_k = u_k*N. Be careful: N does equal m*g since the sticky section is assumed to be on the horizontal part according to the diagram. Given the second y value ( 1.0 m) , solve for the second speed. If the speed is a real number, then it makes it up the ramp; if not, then it does not.

An alternative solution is this: Set the second speed equal to zero and compute the second y value to see if it's equal to 1.0 m.
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For practice, I would recommend you try to re- do this problem using the Work-Energy Theorem before checking back here for more hints.

51.

(a) The angle between the tension force and motion is 18 degrees. For example W_T = T*cos18*D is the work by the tension force, where H = Dsin30 gives is the relation between the vertical height H and the given distance D along the incline. In addition, the gravity work W_g = -mgH, negative since the create is rising. The work by the normal force is zero . Why ?
(b) Compute the Heat = f_k*D, where f_k = u_k*N. The tricky part is getting N. N does not equal mgcos30 because the applied tension force has a component perpendicular to the inclined surface, reducing N below mgcos30. So be careful .

55. Conservation of energy :
½mv₁² + mgy₁+ ½kx₁² + W_EXT =   ½mv₂² + + mgy₂+ ½kx₂²   + HEAT,

where mgy = U_{g ,} ½kx²= U_s , W_EXT is the external force work and HEAT = - W_fk .   Since the friction force is constant,   the HEAT = +f_k*D since W_fk = - f_k*D.

Try doing part (c) first: The external work is zero; the normal force work is zero since the motion is perpendicular to that force. In this case, the first speed is zero, and the first y-value is is 5.0 m. Heat = +f_k*D = u_k*N*D, where D is given. The second y value is zero. The first x value is zero; the second x-value is unknown. Solve for the second x-value.

(a)
½mv₁² + mgy₁ =   ½mv_bot² + + mgy_bot. The first speed is zero and the first y-value is given. Solve for the bottom (bot) speed. The bottom (bot) y-value is zero.

(b) ½mv₁² + mgy₁ =   ½mv_f² + + mgy_f + HEAT. The first speed is zero and the first y-value is given. Solve for the "final"   (f) speed just before   the mass impacts the spring.   Note: The "final" (f) y-value is zero, equal to the bottom y value. Also, HEAT =    +f_k*D = u_k*N*D, where D is given.

(d) This is easy. Set ½mv₁² + mgy₁= HEAT_tot , where HEAT_tot is the total heat after all crossings. The first speed is zero and the first y-value is given. Solve for the total heat, then divide this value by u_k*N. This will give you the total effective distance moved back and forth across the rough part. Finally , divide by 2.0 m to get the total number of crossings.

56.

(a) Use conservation of energy: ½mv₁² + ½kx₁² + mgy₁ =   ½mv₂² + ½kx₂² + mgy₂. For this first leg of the trip, the two heights are equal, so the gravitational potential energy terms cancel. Here 1 refers to the initial position when the spring is compressed and 2 just after the person leaves the spring.
½mv₁² + ½kx₁² =   ½mv₂² + ½kx₂² , where x₁ = -0.50 m and   x₂= 0. The first speed is zero; find the second speed.
(b) Use conservation of energy: ½mv₁² + ½kx₁² + mgy₁ =   ½mv₃² + ½kx₃² + mgy₃+ Heat.    Here 1 refers to the initial position when the spring is compressed and 3 refers to the final position at the maximum height on the opposite slope when the person come momentarily to rest. The first speed is zero and y₁ = 10 m. The final speed is zero and obviously x₃= 0 since the person has left the spring. Now Heat = + f_kD, where D = distance person moves up the opposite incline. Remember Heat is the negative of the work of friction. Now y₃ = Dsin30 from the geometry of the problem. Note f_k = µ_kN, where N = mgcos30. Thus:
½kx₁² + mgy₁ = mgy₃+ Heat. Solve for D and y₃. Note the final height will be larger that 10 m because of the transformation of spring potential energy .

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