Work and Kinetic Energy, Work Done by the Gravitational Force?

Question

In the Figure, a horizontal force Fa of magnitude 22 N is applied to a 2.9 kg book as the book slides a distance d = 0.6 m up a frictionless ramp at angle θ = 30°. The book begins with zero kinetic energy. What is its speed at the end of the displacement?

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Answer 1

The book has two forces working on it- the horizontal force Fa and the gravitational force Ga.

The component of Fa along the incline is Fa cos(30)
=0.866 Fa
The component of Ga along the incline is Ga cos(180-60)
= -Gacos 60 = -0.5 Ga

Ga = mass x accelaration due to gravity =2.9(9.8) =28.42 N
Net force on the book = .866Fa -.5Ga =.866(22)-.5(28.42)
=19.052-14.21
=4.842.
Net accelaration along the incline = 4.842/2.9 =1.67 m/sec^2

v^2 = 2ad = 2(1.67)(0.6) = 2.004 (m/sec)^2

So the kinetic energy = 1/2 mv^2 = 1/2(2.9)2.004 =2.906N-m
Final velocity = sqrt(2.004)= 1.416 m/sec

Answer 2

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Answer 3

Let's write the force equation

f= Fx cos(θ) -mg sin(θ) where

Fx cos(θ) - force along th inclined due to the horizontal force Fa

mg sin(θ) - force due to book's weight that is apposing the motion

Since f=ma then

a=f/m

we know that v=at and S=0.5 a t^2
so t=sqrt(2S/a)
finally
v=at= a sqrt(2S/a) or
v=sqrt(2Sa)
and let's recall that

a= [Fx cos(θ) - mg sin(θ)] /m
a=[22 cos(30) - 2.9 x 9.81 sin(30)]/2.9=1.66 m/s^2

So since

v=sqrt(2Sa)

v=sqrt(2 x 0.6 x 1.66)=1.41 m/s