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Please explain and show all work. Thanks. A 6.6 kg body is at rest on a friction

ID: 2140929 • Letter: P

Question


Please explain and show all work. Thanks.

A 6.6 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F rightarrow acting in the positive direction of an x axis along the track is applied to the body. A stroboscope graph of the position of the body as it slides to the right is shown in the figure. The force F rightarrow is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F rightarrow between t = 0.62 and t = 1.9 s?

Explanation / Answer

Work done = F . r

F = ma

a = acceleration

m= mass

r = distance

so lets find a,

r = u*t + 0.5 * a * t^2

u = initial velocity

t = time

0.04 = 0 * 0.5 + 0.5* a * 0.5^2

a = 0.32


so now lets find r at t=0.62 and t =1.9

r1= 0 * 0.62 + 0.5 * 0.32 * 0.62^2

= 0.061504

r2 = 0 * 1.9 + 0.5 * 0.32 * 1.9^2

= 0.5776


work = ma * (r2 - r1)

= 6.6 * 0.32 (0.5776 - 0.061504)

= 1.08999 N