# work done on the system example

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively, The total work done as the sum of the work done by each force is then seen to be, $W_{total} = W_{gr} + W_N + W_{app} + W_{fr} = 92.0 \, J.$. Given: q = + 4000 kJ (Heat absorbed by sample), ΔV = 0 (Volume remains the same), Work done in the process is given by   W = – Pext × ΔV = – Pext × 0 = 0, Given: q = + 4000 kJ (Heat absorbed by sample), W = + 2000 kJ (Work done by surroundings), ∴ Δ U = + 4000 kJ   + 2000 kJ = + 6000 kJ, Given: q = + 4000 kJ (Heat absorbed by sample), W = – 600 kJ (Work done on the surroundings), ∴ Δ U = + 4000 kJ   –   600 kJ = + 3400 kJ. a) If volume remains constant, what is ΔU? A mass of 10 kg is at a point A on a table. To reduce the kinetic energy of the package to zero, the work $$W_{fr}$$ by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. How much pressure volume work is done and what is the value of ΔU for the reaction of 7.0 g of CO at 1 atm pressure, if the volume change is – 2.8 L. 2CO(g)  +  O2(g) → 2CO2(g)      Enthalpy change = Δ H = – 566 kJ, W =  – Pext × ΔV  = – 1 atm  ×( -2.8) L = 2.8 L atm =  2.8 L atm  × 101.3 J L-1atm-1 = 283.6 J. What are the change in internal energy and enthalpy change of the system? Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Thus, $d' = -\dfrac{W_{fr}}{f} = \dfrac{-95.75 \, J}{5.00 \, N},$. The answers depend on the situation. Example – 05: Calculate the work done in the following reaction when 1 mol of SO 2 is oxidised at constant pressure at 5o °C. The translational kinetic energy of an object of mass $$m$$ moving at speed $$v$$ is $$KE = \frac{1}{2}mv^2$$. is done. Thus, as expected, the net force is parallel to the displacement, so that $$\theta = 0$$ and $$cos \, \theta = 1$$, and the net work is given by, The effect of the net force $$F_{net}$$ is to accelerate the package from $$v_0$$ to $$v$$ The kinetic energy of the package increases, indicating that the net work done on the system is positive. negative work The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3. Ans: Work done by the surroundings on the system in the reaction is 1758 J. CO reacts with O2 according to the following reaction. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. Does it remain in the system or move on? Use work and energy considerations. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done is e $$(F \, cos \, \theta)_{i(ave)}d_i$$ for each strip, and the total work done is the sum of the $$W_i$$. In SI system unit of work is 1Nm and is given a name Joule(J). Because the mass $$m$$ and the speed $$v$$ are given, the kinetic energy can be calculated from its definition as given in the equation $$KE = \frac{1}{2}mv^2$$. The horizontal friction force is then the net force, and it acts opposite to the displacement, so $$\theta = 180^o$$. Where is the energy you spend going? The net work equals the sum of the work done by each individual force. This expression is called the work-energy theorem, and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. We will now consider a series of examples to illustrate various aspects of work and energy. Given:  Initial volume = V1 = 6 dm³ = 6 × 10-3 m³, Final volume = V2 = 16 dm³ = 16 × 10-3 m³, Pext = 2.026 x 105 Nm-2, ΔU = 418 J. The initial and the final points of the path of the object lie on the same horizontal line. The person actually does more work than this, because friction opposes the motion. is the energy associated with translational motion. Login to view more pages. What is its kinetic energy? Sign convention: work done by a system is positive, and the work done on a system is negative. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work. Example $$\PageIndex{2}$$: Determining the Work to Accelerate a Package. Does it remain in the system or move on? In this section we begin the study of various types of work and forms of energy. NCERT Question 2 - ΔH = qp = Heat supplied at constant pressure = + 6 kJ, Ans: The change in internal energy is 4.5 kJ and enthalpy change is 6 kJ. We know that once the person stops pushing, friction will bring the package to rest. Suppose that you push on the 30.0-kg package in Figure 7.03.2. with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N. (a) Calculate the net work done on the package. NCERT Question 8 - State whether work is on the system or by the system. Calculate pressure-volume type work and ΔU. State whether work is on the system or by the system. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. For example, if the lawn mower in [link](a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. Enthalpy change = ΔH =? What is the work done by the force of gravity on the object? Subscribe to our Youtube Channel - https://you.tube/teachoo. The net force arises solely from the horizontal applied force $$F_{app}$$ and the horizontal friction force $$f$$. Such a situation occurs for the package on the roller belt conveyor system shown in Figure. Figure (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an $$F \, cos \, \theta$$ vs. $$d$$ graph. Find Enthalpy change if ΔU is 418 J. The theorem implies that the net work on a system equals the change in the quantity $$\frac{1}{2}mv^2$$. The calculated total work $$W_{total}$$ as the sum of the work by each force agrees, as expected, with the work $$W_{net}$$ done by the net force. Solving for acceleration gives $$a = \frac{v^2 - v_0^2}{2d}.$$ When $$a$$ is substituted into the preceding expression for $$W_{net}$$ we obtain, $W_{net} = m \left(\dfrac{v^2 - v_0^2}{2d} \right)d.$, The $$d$$ cancels, and we rearrange this to obtain, \[W_{net} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. ### No Comments

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