<p><span style=\"font-size: medium;\">An 8.70-kg block slides with an initial sp

Question

<p><span>An 8.70-kg block slides with an initial speed of 1.60 m/s down a ramp inclined at an angle of 27.1degrees with the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.85.</span></p>
<p><span><span>Use energy conservation to find the distance the block slides before coming to rest.</span></span></p>
<p><span><span>Can you please show all steps when deriving the answer? the answer is d= .43 m but I do not know how to obtain it</span></span></p>

Explanation / Answer

From the Work-Energy Theroem, the block will have KE and PE at the top that friction will take away

The work by friction = Ffd

Ffd = mgcosd

The PE change will be mgh   (h = dsin)

The change in KE will be .5mv2

The equation becomes

mgcosd = mgdsin + .5mv2    (m cancels)

Simplify and factor d

d(gcos - gsin) = .5v2

d[(.85)(9.8)(cos 27.1) - (9.8)(sin 27.1)] = .5(1.6)2

d = .434 m

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