A block of mass m = 0.86 kg slithers down an icy slope, as shown in the figure.

Question

A block of mass m = 0.86 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? What is the frictional force on the block, if the coefficient of friction between the block and the ice is 0.419? 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

u = mgh = 0.86*9.8*12 = 101.1 J

f = mue*N = 0.419*0.8*0.86*9.8 = 2.825 N

wf = -f*s = -2.825*20 = -56.5 J

w tot = wg+wf = 101.1-56.5 = 44.6 J

w = 1/2mv^2

k = 1/2mv^2 = 44.6 J

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