document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=157529>Question 157529</A>: <I><SPAN STYLE=\"word-break: break-all;\"> With a force of 15 N, a person is pushing a 30 kg crate over a horizontal surface. Suppose the crate is moving at a constant velocity.\r<BR>\n" );
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document.write( "	a. What is the force of friction?\r<BR>\n" );
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document.write( "	b. What is the acceleration of the crate?\r<BR>\n" );
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document.write( "	c. Suppose the person lets go of the crate. What is the new acceleration that initially occurs right after the crate is let go? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #116088 by </B><A HREF=/tutors/aboutme.mpl?userid=Fombitz><B>Fombitz(32388)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fombitz><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Fombitz><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=116088\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A)15 N. You are precisely balancing the frictional force with your push.<BR>\n" );
document.write( "B)Acceleration is zero because velocity is constant.<BR>\n" );
document.write( "C)Without you providing the 15N in the forward direction, an immediate force of 15N, from dynamic friction, will arise in the reverse direction (opposite to the direction of motion).<BR>\n" );
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