Posted by Bob on Tuesday, May 8, 2012 at 3:35pm.
(a)
Summing the components of the forces along the vertical direction provides
F• sinα + N - m•g = 0,
=> F• sinα = m•g - N.
For the horizontal direction we have
F•cosα – F(fr) = 0
F(fr) = k•N, => F•cosα = k•N,
F• sinα/ F•cosα = (m•g – N)/k•N,
tan α = (m•g – N)/k•N,
N = m•g/(k•tan +1) = 25.2•9.8/(0.4•tan21.6º +1) = 213 N.
(b)
If F•cosα = k•N,
F = k•N/cosα.
The work done on the cart is given by
W(fr) =F•s•cosα = k•N•s•cos α/cosα = =k•N•s = 0.4•213•13.7 = 1167 J =
=1.167 kJ.
(c)
The energy lost due to friction is just the work done by friction on the cart:
W(fr) = F(fr)•s•cos 180º = k•N•s•(-1) =
= - k•N•s = - 0.4•213•13.7 =
= - 1167 J = - 1.167 kJ.
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