Work Done by Friction on 17.7 kg Block: Find Magnitude

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A 17.7 kg block is being dragged over a rough surface by a 187 N force at a 33.6-degree angle, with a coefficient of kinetic friction of 0.199. The work done by the applied force was calculated correctly, but the work done by friction caused confusion due to the negative direction of friction. The correct approach involves using the normal force adjusted for the applied force's vertical component, leading to a calculation of work done by friction as W(fric) = - (mu)(normal force)(d). After recalculating, the correct magnitude of work done by friction was found to be approximately -647.52 J, highlighting the importance of accurate arithmetic in solving physics problems. The discussion emphasizes the need for careful calculations when determining work done against friction.
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1. A 17.7 kg block is dragged over a rough, hor-
izontal surface by a constant force of 187 N
acting at an angle of angle 33.6 above the
horizontal. The block is displaced 46.5 m and
the coefficient of kinetic friction is 0.199. Find the magnitude Work done by the force of friction



2.W=F*D
W=(mu)(mg)cos(theta)



3. I have already correctly solved the work done by the applied force at angle 33.6 above the horizontal via (187N)(46.5m)(cos(33.6)) = 7240 (sig figs). However the next question regarding the work done by the force of friction for some reason has me quite ticked off. I solve basically the same way however the friction acts opposite the force of motion makings its value negative. the question specifically calls for its magnitude not its direction so took its absolute value. W=F*D yeilds F=(muk)(mg) where mg is the normal force. making work W=(mu)(m)(g)cos(180). my thoughts on the angle is that the only component of the applied force the force of friction acts opposite to is the, in my axis, the positive x direction. opposite of 0 degrees is 180 causing the value to be negative. the answer is incorrect. Solving the same equation minus the cos(theta) is wrong. :(

edit: i used a free body diagram
 
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ive also thought it through like this:

W(fric) = -f(fric) d = - (mu) (normal force) (d)
= -(mu)(mg - F sin@)(d)
W(fric) = - (0.199)(17.7 kg x 9.8 m/s/s - (187N)(sin 33.6) ) (46.5m)
= -(0.199)(69.976)(46.5)
= -2599.845 J

but is incorrect.
 
I got-
mu(mgcos(33.6)) * 46.5
 
Idividebyzero said:
ive also thought it through like this:

W(fric) = -f(fric) d = - (mu) (normal force) (d)
= -(mu)(mg - F sin@)(d)
W(fric) = - (0.199)(17.7 kg x 9.8 m/s/s - (187N)(sin 33.6) ) (46.5m)
= -(0.199)(69.976)(46.5)
= -2599.845 J

but is incorrect.
This looks like a correct method.

I get that -(0.199)(69.976)(46.5) = -647.52 J

Check your arithmetic.
 
SammyS said:
This looks like a correct method.

I get that -(0.199)(69.976)(46.5) = -647.52 J

Check your arithmetic.

thanks a bunch it was the arithmetic that killed me. i calculated the net work on the object as it was the next question using the 647.xx as the frictional work and the answer was correct. i made a mistake in my math... now to find where
 

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