1.
work done is given by,
W = F*d*cos(A)
here, F = force = friction force
F = *N
d = distance travelled = 1 m
A = angle between F & d = 180 deg
where, = 0.5
N = normal force = m*(g*cos)
given = inclined angle = 30 deg
m = mass = 1 kg
So, N = 1*9.8*cos(30 deg) = 8.5
then, W = 0.5*8.5*1*cos(180 deg)
W = -4.25 J
therefore correct option is B. Approximately -4.2 J
2.
Center of mass is given by:
In x-direction
Xcm = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3)
In y-direction
Ycm = (m1y1 + m2y2 + m3y3)/(m1 + m2 + m3)
m1 = 2 kg, (x1, y1) = (0,0)
m2 = 2 kg, (x2, y2) = (1, 0)
m3 = 2 kg, (x3, y3) = (0, 1)
m4 = 4 kg, (x4, y4) = (1, 1)
Now
Xcm = (2*0 + 2*1 + 2*0 + 4*1)/(2+2+2+4) = 0.6 m
Ycm = (2*0 + 2*0 + 2*1 + 4*1)/(2+2+2+4) = 0.6 m
Center of mass will be at (Xcm, Ycm) = (0.6, 0.6) m
So, magnitude of distance of center of mass from origin = d = sqrt(0.6^2 + 0.6^2)
d = 0.8485 m = approx. 0.85 m
direction of center of mass = = arctan(Ycm/Xcm) = arctan(1/1)
= 45 deg from X axis direction
therefore correct option is B.
Magnitude : approx. 0.85 m
Direction : 45 deg from X axis direction
Please upvote.
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