A horizontal force Fa magnitude 20.0 N is applied to a 2.99 kg book as the book slides a distance d = 0.93 m up a frictionless ramp at angle θ = 30°. (a) During the displacement, what is the net work done on the book by Fa , the gravitational force on the book, and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is its speed at the end of the displacement?
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2015-11-09 8:15 am
回答 (1)
2015-11-09 10:58 am
(a) During the displacement, what is the net work done on the book by
Fa, the gravitational force on the book, and the normal force on the
book?
E = Fa*L = 20 * 0.93 = 18.6 J
G = mg * sin30 = 2.99 * 9.8 * 0.5 = 14.651 N
N = mg * cos30 = 2.99 * 9.8 * √3 / 2 = 14.651√3 N
(b) If the book has zero kinetic energy at the start of the
displacement, what is its speed at the end of the displacement?
a = (Fa - G)/m = (20 - 14.651)/2.99 = 1.78896 m/s^2
V = √(2aL) = √(2*1.78896*0.53) = 1.377 m/s
Fa, the gravitational force on the book, and the normal force on the
book?
E = Fa*L = 20 * 0.93 = 18.6 J
G = mg * sin30 = 2.99 * 9.8 * 0.5 = 14.651 N
N = mg * cos30 = 2.99 * 9.8 * √3 / 2 = 14.651√3 N
(b) If the book has zero kinetic energy at the start of the
displacement, what is its speed at the end of the displacement?
a = (Fa - G)/m = (20 - 14.651)/2.99 = 1.78896 m/s^2
V = √(2aL) = √(2*1.78896*0.53) = 1.377 m/s
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