Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62° north of west. What is the magnitude of the body's acceleration?
Answer in the book (with no workings) says 2.9m/s2
I get ~1.7m/s2
F(net,x) = ma
F1(x) = 9.0N
F2(x) = -8.0cos(62) = ~ -3.8N
F(net,x) = F1(x) + F2(x) = 9.0N - 3.8N = 5.2N
5.2N = 3.0kg * a(m/s2)
a = 5.2N / 3.0kg = ~1.7m/s2
Seems like a straight forward question, what am I doing wrong?
+36916 the forces acting laterally are
+9 N to the east
and -8cos62 = -3.755 N
Net force is +5.244 east f= ma 5.244 = 3 kg a a = 1.75 m/s^2 east
Either the book has the wrong answer or the wrong info in the question!
Good job !
+36916 ..as an 'aside' if the angle is 6.2 degrees north of west, the answer comes out to a = 2.87 m/s^2
Maybe there is a typo in the angle of the 8N force ??
