# How does work enter into this law of conservation of energy problem?

A 10kg block slides down a hill that is 200m tall as shown in the figure below with an initial speed of 12m/sec.

a) What is the speed of the block when it reaches the bottom of the hill at position B assuming no friction?

b) What is the final speed of the block at position B after traveling a total distance of 500m given a coefficient of kinetic friction of 0.21 between the block and the surface?

Sol:

At position A, block has potential energy because it is above position B. It has capability of falling. Because the block is in motion, It has kinetic energy. At position B, it only has kinetic energy.

At position A, there is kinetic energy and potential energy. At position B, it only has kinetic energy. $$PE + KE = KE$$ All of the potential energy at position A is going to kinetic energy of the object. The object speed will increase beyond 12m/sec.

$$mgh + \large mv_{0}^{2} = \large mv_{f}^{2}$$

$$2gh + V_{0}^{2} = V_{f}^{2}$$

$$V_{f}^{2} = 12^{2} + 2(9.8)(200)$$

$$V_{f} = 63.75 m/sec$$

b) let consider total distance of 500m is horizontal distance of inclination.

At position A, there is kinetic energy and potential energy. There is friction, so final speed will be less than 63.75 m/sec. $$PE + KE = KE$$ sum of the energy from left side is not all go into kinetic energy. some of it go into work done by friction. If work done by friction put in left side then it is going to be $$-W$$; friction is going to be decrease total energy of the object. Then it is positive $$W$$ on right side.

$$PE + KE = KE + W$$

Friction's job is to take mechanical energy from the object.

$$→W = f_{k}.d$$

$$→W = μ_{k}.F_{N}.d$$

$$→W = μ_{k}.mg.d$$

$$→ W = 0.21(10)(9.8)(500) = 10,290J$$

$$PE + KE = KE + W$$

$$→ mgh + \large\frac{1}{2}\normalsize mv_{0}^{2} = \large\frac{1}{2}\normalsize mv_{f}^{2} + W$$

$$→ 10(9.8)(200) + \large\frac{1}{2}\normalsize(10)(12) = KE_{F} + 10290 J$$

$$→ 19600 + 720 = KE_{F} + 10290$$

$$→ 10,030 = KE_{F}$$

$$→\large\frac{1}{2}\normalsize mv_{f}^{2} = 10,030$$

$$→ v_{f}^{2} = 2006$$

$$→ v_{f} = 44.8 m/sec$$

How does work enter into this law of conservation of energy problem?

• For part b, the problem statement is not clear if 500 meters is the total distance along the surface of the hill, or the horizontal distance between the starting and final position. Are you saying 500 m is the horizontal distance or is the problem statement saying that? – Bob D Jul 8 '19 at 13:30

How work interferes into law of conservation of energy problem?

Work does not interfere with the law of conservation of energy. The final kinetic energy in part (b) is less than the final kinetic energy in part (a), because in part (b) some kinetic energy is converted to heat due to the friction work.

Without friction the final kinetic energy of part (a) using your velocity is: (.5)(10)(63.76)$$^2$$=20,320 J

With friction the final kinetic energy of part (b), using your velocity is:(0.5)(10)(44.8)$$^2$$=10,035 J

The difference in kinetic energy is:20,320-10,035 = 10,285 J

Which is approximately equal to the heat dissipated due to the friction work you calculated of 10,290 J (the difference probably due to round offs of your calculated velocities) based on the assumption that 500 m is the horizontal distance traveled..

Hope this helps.