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A 1.5-kg physics textbook is initially at rest on a steel table. The textbook is then
pushed with a constant force of 5.0 N. Friction with a magnitude of 3.0 N is
exerted on the moving book by the surface of the table. Determine the final
velocity of the textbook after it has been pushed 0.50 meter across the table.

Answer :

indrawboy

∑F = m.a

5 - 3 = 1.5 x a

a = 1.3 m/s²

vf² = vi² + 2ad (vi = o at rest)

vf² = 2 x 1.3 m/s² x 0.5

vf² = 1.3

vf = 1.14 m/s

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