A 2.9 kg body is at rest on a frictionless horizontal air track when a horizonta

Question

A 2.9 kg body is at rest on a frictionless horizontal air track when a horizontal force F acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in the figure. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. a) How much work is done on the body by the applied force F between t = 0 and t = 0.5 s? (HINT: Find the average acceleration over t = 0 to t = 0.5 s.)

Explanation / Answer

Given: s=0.2m based on the graph. t=.5s u=0 Solve for acceleration, a, using the formula: s=ut+1/2at^2 where s= the distance traveled u=the initial velocity a=acceleration t=time Substitute given values: 0.2=0*0.5+1/2a*0.5^2 0.2=1/2a a=0.75m/s^2 Now solve for F using the formula: F=Ma where F is the applied force, M the mass, and a the acceleration. Given: M=2.9kg a=0.75m/s^2 Substitute given values: F=2.9*0.75 =2.175N Solve for the work using the formula: Work=F*s where F is the applied force and s is the distance traveled. Given: F=2.175N s=0.2m Substitute given values: Work =2.175*0.2 =0..35 joule

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