A horizontal force of magnitude 33.2 N pushes a block of mass 3.94 kg across a f
ID: 1429327 • Letter: A
Question
A horizontal force of magnitude 33.2 N pushes a block of mass 3.94 kg across a floor where the coefficient of kinetic friction is 0.629. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 3.48 m across the floor? (b) During that displacement, the thermal energy of the block increases by 44.5 J. What is the increase in thermal energy of the floor?(c) What is the increase in the kinetic energy of the block?
(a) Number Units This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3timesExplanation / Answer
a) Work done by a force = F.d = |F||d| cos@
where @ is the angle between force vector and displacement.
W = 33.2 x 3.48 x cos0 = 130.81 J
b) friction force = uN = u mg = 0.629 x 3.94 x 9.8 = 24.29 N
Work done by friction = f d cos180 = - 84.52 J
this energy converts into thermal energy.
so increase in thermal enery of floor = 84.52 - 44.5 =40 J
c) using work energy theorem.
Work done by force + work done by friction = change in KE
130.81 - 84.52 = 3.94 ( v^2 - 0 ) /2
v = 4.85 m/s
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