Figure 6.4-1: Work-Energy Apparatus. Consider an object that moves only along th

Question

Figure 6.4-1: Work-Energy Apparatus. Consider an object that moves only along the positive X axis that is subjected to a constant force pointing in the positive X direction. Show the work done by this force is the area under a Force vs. Displacement plot. A cart of mass (851.8 + - 0.1) g is configured in a setup as shown in Figure 6.4-1. The tension in the string is 0.36 N and the cart has a displacement of 40.9 cm. Determine the work done by the tension. The cart in the previous example has a maximum velocity of (0.580 + - 0.017) m/s. Determine the kinetic energy of the cart and compare the result with the work done on the cart in Problem 2 to confirm the Work-Energy Theorem.

Explanation / Answer

a) SInce, the force is constant (magnitude as well as direction).

the force versus displacement graph will be horizontal .

At any instance teh area under the graph wll be F*D [

Here, F is the force .

D, is the displacement .

ALso for this kind of force the Total work done is -> w = Force. Displacement [Dot product]

W = F *D [since displacement will be in the direction of the force ]

W =Area under the graph [hence proved]

b) Tension = 0.36 N

displacement = 40.9 cm = 0.409m

Work done = T*D = 0.36 *0.409 J = 0.14724 J [ANS]

c) Maximum velocity(v) = 0.58 +0.017% m/s

MAss of the cart (m) = 851.8 +0.1% g = 0.8518 +0.1% kg

Kinetic energy = 0.5m*v*v = 0.5*0.8518 * 0.58* 0.58 + 0.134% J = 0.1432 + 0.134% J

Kinetic enrgy = 0.1432 + 0.134% which is almost equal to 0.14724 J [Work done]

Hence proved.

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