A 15.2 kg block is dragged over a rough, hor- izontal surface by a constant forc

Question

A 15.2 kg block is dragged over a rough, hor- izontal surface by a constant force of 146 N acting at an angle of angle 27 above the horizontal. The block is displaced 18 m and the coefficient of kinetic friction is 0.193.
A. Find the work done by the 146 N force. The acceleration of gravity is 9.8 m/s2 . Answer in units of J.
B. Find the magnitude of the work done by the force of friction. Answer in units of J. Find the magnitude of the work done by the force of friction. Answer in units of J.
C. What is the sign of the work done by the frictional force? (Positive, negative, or zero?)

D. Find the work done by the normal force. Answer in units of J.

E. What is the net work done on the block? Answer in units of J. What is the net work done on the block? Answer in units of J.
146 N 15.2 kg = 0.193

Explanation / Answer

a)

Work done by force = F*cos(27)*d

= 146*cos(27 deg)*18

= 2341.6 J

b)

Work done by friction = -0.193*(mg - F*sin(27))*d

= -0.193*(15.2*9.8 - 146*sin(27 deg))*18

= -287.2 J

c)

Sign is negative <---- answer

It is due to the fact that friction acts in the opposite direction to the direction of didsplacement

d)

Work done by normal force W = 0 <--- as displacement is perpendicular to the normal force

e)

Net work done = 2341.6 - 287.2 = 2054.4 J

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