A block of mass m = 1.06 kg slithers down an icy slope, as shown in the figure.
ID: 1616727 • Letter: A
Question
A block of mass m = 1.06 kg slithers down an icy slope, as shown in the figure. The slant distance the block slides by is 20 m, during which it descends by a distance 12 m, and moves a horizontal distance (by Pythagoras' theorem) of 16 m. Resolving its weight into components, the normal component of the gravitational force is N = 0.8 m g and the tangential component is T = 0.6 m g. How much does the potential energy decrease from the top to the bottom of the slope? J What is the frictional force on the block, if the coefficient of friction between the block and the ice is 0.399? N What is the work done against the friction force by the time the block reaches the bottom of the slope? J Ignoring all forces on the block apart from gravity and friction, what is its kinetic energy at the bottom of the slope? J
Explanation / Answer
a)
dU=U2-U1
=m*g*h2-m*g*h1
=m*g*h2-0
=1.06*9.8*12-0
=124.66 J
b)
frictional force, fk=uk*N
=0.399*0.8*m*g
=0.399*0.8*1.06*9.8
=3.31 N
c)
work done, w_fric=fk*d
=3.31*20
=66.2 J
d)
by using energy relation,
K2-K1=U1-U2-W_fric
==>
K1-K2=U2-U1+W_fric
K1-0=124.66-0+66.2
K1=190.86
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.