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Unformatted text preview: Name: __________________________________ Physics 1112, Fall 2008, Assignment 9 Solutions Cornell University 1 Physics 1112, Fall 2008, Assignment 9 Solutions 1. (Young & Freedman (Y & F) 7.74) See textbook Figure 7.43. kg 00 . 2 = m . = 1 . 53 . N/m 120 = k . 40 . s = . 20 . k = . m/s 00 . i = v . The initial distance from the package to the spring is m 00 . 4 = d . Since the package is released from rest, we need to see whether the static friction force can be large enough to prevent it from sliding down the incline. p on i s s p, on i N f . So, we need to calculate p on i N . Apply Newtons 2 nd law along the inc y direction: ) cos( p on E p on i = W N . ) cos( ) cos( p on E p on i mg W N = = . Now apply Newtons 2 nd law along the inc x direction: inc ) sin( p on E p on i x ma W f = . So, if s p, on i f can equal ) sin( ) sin( p on E mg W = then the package will remain at rest at the top of the incline. ) cos( ) sin( s mg mg . s ) tan( . 40 . 33 . 1 ) 1 . 53 tan( ) tan( s = > = = , so the package does indeed slide down the incline towards the spring. Note that as the package slides along the incline, the normal force p on i N r never does any work on it because p on i N r is always perpendicular to the displacement of the package. (a) The kinetic friction force k p, on i f r is a nonconservative force and it always points in the opposite direction to the displacement of the package. The work done by this force as the package slides down the incline to the spring is d mg d N d W ) cos( ) ( ) ( k inc inc p on i k inc k p, on i k p, on i = = = i i i f f r r . The initial kinetic and potential energies are: i = K . ) sin( i mgd U = . The kinetic and potential energies just before the package hits the spring are: 2 s 2 1 s mv K = . s = U . s s i i k p, on i U K W U K + = + + f r . 2 s 2 1 k ) cos( ) sin( mv d mg mgd = . )) 1 . 53 cos( ) 20 . ( ) 1 . 53 )(sin( m 00 . 4 )( m/s 81 . 9 ( 2 )) cos( ) (sin( 2 2 k s = = gd v ....
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