A 11.0 kg block is dragged over arough, horizontal surface by a 90.0 Nforce acti

Question

A 11.0 kg block is dragged over arough, horizontal surface by a 90.0 Nforce acting at 20.0° above the horizontal. The block isdisplaced 7.00 m, and the coefficientof kinetic friction is 0.300. (a) Find the work done on the block by the90.0 N force.
J
(b) Find the work done on the block by the normal force.
J
(c) Find the work done on the block by the gravitational force.
J
(d) What is increase in internal energy of the block-surface systemdue to friction?
J
(e) Find the total change in the block's energy.
J (a) Find the work done on the block by the90.0 N force.
J
(b) Find the work done on the block by the normal force.
J
(c) Find the work done on the block by the gravitational force.
J
(d) What is increase in internal energy of the block-surface systemdue to friction?
J
(e) Find the total change in the block's energy.
J

Explanation / Answer

the forces acting on the block are Fx = F2 - k* n = F * cos -k* n -------------(1) Fy = (n + F1) - w = 0 or n = w - F1 = w - F * sin ------------(2) from (1) and (2) Fx = F * cos - k* (w - F *sin) (a)the work done on the block by the 90.0 N force W = Fx * S = [F * cos - k* (w - F* sin)] * S F = 90.0 N = 20.0o w = m * g m = 11.0 kg and g = 9.8 m/s^2 k= 0.300 S = 7.00 m (b)the work done on the block by the normal force W1 = n * S = (w - F * sin) * S (c)the work done on the block by the gravitational force W2 = mg * S (d)the conservation of energy equation,for the block-surfacesystem due to friction K + Eint = 0 or Eint= -K Here,K = W (e)the total change in the block's energy W3 = W - W1 w = m * g m = 11.0 kg and g = 9.8 m/s^2 k= 0.300 S = 7.00 m (b)the work done on the block by the normal force W1 = n * S = (w - F * sin) * S (c)the work done on the block by the gravitational force W2 = mg * S (d)the conservation of energy equation,for the block-surfacesystem due to friction K + Eint = 0 or Eint= -K Here,K = W (e)the total change in the block's energy W3 = W - W1 W3 = W - W1

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