A 1.8 kg block is moved at constant speed over a surface for which coefficient of friction is equal to 1/4 . It is pulled by a force F acting at 45 degree with horizontal. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity?
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a) the component of the force along the direction of displacement is F/sqrt(2)
the work done by this force is F.dx=2F/sqrt(2)=(sqrt(2).F)J=36J (see part c)
b)the static force of friction=umg=1/4*1.8*10=4.5N
now,it is displaced by 2m..so,work=-(4.5*2)= -9J(-ve sign since friction is opp. direction to the displacement..)
c)since there is no displacement in vertical direction, F/sqrt(2)=mg=1.8*10=18N..
also,work done by gravity is 0J..
the work done by this force is F.dx=2F/sqrt(2)=(sqrt(2).F)J=36J (see part c)
b)the static force of friction=umg=1/4*1.8*10=4.5N
now,it is displaced by 2m..so,work=-(4.5*2)= -9J(-ve sign since friction is opp. direction to the displacement..)
c)since there is no displacement in vertical direction, F/sqrt(2)=mg=1.8*10=18N..
also,work done by gravity is 0J..
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(a)Work done by the force F:
W=F s cosθ
=F×2×cos45°
=F√2
F cos45°−(μ)(mg−F sin45°)=0
F cos45°−14(1.8×10−F sin45°)=0
F/√2+F/4√2=18/4
5F/4√2=18/4
F=18√2/5=5.09 N
W=√2×5.09 = 7.19J
(b)Work done by the friction:
W=μ(mg−F sin45°)s cos180°
=1/4 (1.8×10−5.09/√2) × 2 × (−1)
=−1/4 × 14.4 × 2
=−7.2 J
(c)Work done by gravity:
W=mg × scosθ
=mg × scos90°
=0
Hope it helps
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